Integration of irrational functions. Complex integrals
Let's remember our happy school years. Pioneers in mathematics lessons, when starting to study roots, first of all became acquainted with the square root. We will go the same way.
Example 1
Find the indefinite integral
Analyzing the integrand, you come to the sad conclusion that it does not at all resemble table integrals. Now, if all this stuff was in the numerator, it would be simple. Or there would be no root below. Or a polynomial. None fraction integration methods They don't help either. What to do?
The main technique for solving irrational integrals is a change of variable, which will rid us of ALL roots in the integrand.
Note that this replacement is a little peculiar; its technical implementation differs from the “classical” replacement method, which was discussed in the lesson Substitution method in indefinite integral.
In this example, you need to replace x = t 2, that is, instead of the “X” under the root we will have t 2. Why the replacement? Because, and as a result of the replacement, the root will disappear.
If instead of the square root we had in the integrand, then we would have made the substitution. If it had been there, they would have carried it out and so on.
Okay, we will turn into . What happens to the polynomial? There are no difficulties: if , then 
.
It remains to be seen what the differential will turn into. This is done like this:
We take our replacement and hang differentials on both parts:
(we will describe it in as much detail as possible).
The solution format should look something like this:
.
Let's replace: 
.
.
(1) We carry out the substitution after the replacement (how, what and where is already considered).
(2) We take the constant outside the integral. The numerator and denominator are reduced by t.
(3) The resulting integral is tabular; we prepare it for integration by selecting the square.
(4) Integrate over the table using the formula
.
(5) We carry out the reverse replacement. How is this done? We remember why we danced: if, then.
Example 2
Find the indefinite integral
This is an example for you to solve on your own. Full solution and answer at the end of the lesson.
Somehow it turned out that in Examples 1, 2 there is a “bare” numerator with a lonely differential. Let's fix the situation.
Example 3
Find the indefinite integral
A preliminary analysis of the integrand again shows that there is no easy way. And therefore you need to get rid of the root.
Let's replace: .
For we denote the ENTIRE expression under the root. The replacement from the previous examples is not suitable here (more precisely, it can be done, but it will not get rid of the root).
We hang differentials on both parts:
We've sorted out the numerator. What to do with in the denominator?
We take our replacement and express from it: .
If, then.
(1) We carry out the substitution in accordance with the replacement performed.
(2) Comb the numerator. Here I chose not to take the constant out of the integral sign (you can do it this way, it won’t be an error)
(3) We expand the numerator into a sum. Once again, we strongly recommend that you read the first paragraph of the lesson Integrating Some Fractions. There will be plenty of rigmarole with decomposing the numerator into a sum in irrational integrals; it is very important to practice this technique.
(4) Divide the numerator by the denominator term by term.
(5) We use the linearity properties of the indefinite integral. In the second integral we select a square for subsequent integration according to the table.
(6) We integrate according to the table. The first integral is quite simple, in the second we use the tabular formula of the high logarithm 
.
(7) We carry out the reverse replacement. If we carried out a replacement, then, back: .
Example 4
Find the indefinite integral
This is an example for you to solve on your own; if you have not carefully worked through the previous examples, you will make a mistake! Full solution and answer at the end of the lesson.
In principle, integrals with several identical roots, for example
Etc. What to do if the integrand has roots different?
Example 5
Find the indefinite integral
Here comes the reckoning for the bare numerators. When such an integral is encountered, it usually becomes scary. But the fears are in vain; after making a suitable replacement, the integrand simplifies. The task is the following: to carry out a successful replacement in order to immediately get rid of ALL roots.
When given different roots, it is convenient to adhere to a certain solution scheme.
First, we write out the integrand function on a draft, and present all the roots in the form:
We will be interested denominators degrees:
This section will discuss the method of integrating rational functions. 7.1. Brief information about rational functions The simplest rational function is a polynomial of the tith degree, i.e. a function of the form where are real constants, and a0 Ф 0. The polynomial Qn(x) whose coefficient a0 = 1 is called reduced. A real number b is called the root of the polynomial Qn(z) if Q„(b) = 0. It is known that each polynomial Qn(x) with real coefficients is uniquely decomposed into real factors of the form where p, q are real coefficients, and the quadratic factors have no real roots and, therefore, cannot be decomposable into real linear factors. By combining identical factors (if any) and assuming, for simplicity, that the polynomial Qn(x) is reduced, we can write its factorization in the form where are natural numbers. Since the degree of the polynomial Qn(x) is equal to n, then the sum of all exponents a, /3,..., A, added to the double sum of all exponents ω,..., q, is equal to n: The root a of a polynomial is called simple or single , if a = 1, and multiple if a > 1; the number a is called the multiplicity of the root a. The same applies to other roots of the polynomial. A rational function f(x) or a rational fraction is the ratio of two polynomials, and it is assumed that the polynomials Pm(x) and Qn(x) do not have common factors. A rational fraction is called proper if the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. If m n, then the rational fraction is called an improper fraction, and in this case, dividing the numerator by the denominator according to the rule for dividing polynomials, it can be represented in the form where are some polynomials, and ^^ is a proper rational fraction. Example 1. A rational fraction is an improper fraction. Dividing by a “corner”, we have Therefore. Here. and it's a proper fraction. Definition. The simplest (or elementary) fractions are rational fractions of the following four types: where are real numbers, k is a natural number greater than or equal to 2, and the square trinomial x2 + px + q has no real roots, so -2 _2 is its discriminant In algebra the following theorem is proved. Theorem 3. A proper rational fraction with real coefficients, the denominator of which Qn(x) has the form decomposes in a unique way into the sum of simple fractions according to the rule Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions First Euler substitution Second Euler substitution Third Euler's substitution In this expansion there are some real constants, some of which may be equal to zero. To find these constants, the right-hand side of equality (I) is brought to a common denominator, and then the coefficients at the same powers of x in the numerators of the left and right sides are equated. This gives a system of linear equations from which the required constants are found. . This method of finding unknown constants is called the method of undetermined coefficients. Sometimes it is more convenient to use another method of finding unknown constants, which consists in the fact that after equating the numerators, an identity is obtained with respect to x, in which the argument x is given some values, for example, the values of the roots, resulting in equations for finding the constants. It is especially convenient if the denominator Q„(x) has only real simple roots. Example 2. Decompose the rational fraction into simpler fractions. This fraction is proper. We decompose the denominator into multiplies: Since the roots of the denominator are real and different, then, based on formula (1), the decomposition of the fraction into the simplest will have the form: Reducing the right honor “of that equality to the common denominator and equating the numerators on its left and right sides, we obtain the identity or We find unknown coefficients A. 2?, C in two ways. First way Equating the coefficients for the same powers of x, t.v. with (free term), and the left and right sides of the identity, we obtain a linear system of equations for finding the unknown coefficients A, B, C: This system has a unique solution C The second method. Since the roots of the denominator are torn at i 0, we get 2 = 2A, whence A * 1; g i 1, we get -1 * -B, from which 5 * 1; x i 2, we get 2 = 2C. whence C» 1, and the required expansion has the form 3. Rehlozhnt not the simplest fractions rational fraction 4 We decompose the polynomial, which is in the opposite direction, into factors: . The denominator has two different real roots: x\ = 0 multiplicity of multiplicity 3. Therefore, the decomposition of this fraction is not the simplest: Reducing the right-hand side to a common denominator, we find or The first method. Equating the coefficients for the same powers of x in the left and right sides of the last identity. we obtain a linear system of equations. This system has a unique solution and the required expansion will be the Second method. In the resulting identity, putting x = 0, we obtain 1 a A2, or A2 = 1; field* gay x = -1, we get -3 i B), or Bj i -3. When substituting the found values of the coefficients A\ and B) and the identity will take the form or Putting x = 0, and then x = -I. we find that = 0, B2 = 0 and. this means B\ = 0. Thus, we again obtain Example 4. Expand the rational fraction 4 into simpler fractions. The denominator of the fraction has no real roots, since the function x2 + 1 does not vanish for any real values of x. Therefore, the decomposition into simple fractions should have the form From here we get or. Equating the coefficients of the synax powers of x in the left and right sides of the last equality, we will have where we find and, therefore, It should be noted that in some cases expansions into simple fractions can be obtained faster and easier by acting in some other way, without using the method of indefinite coefficients For example, to obtain the decomposition of the fraction in example 3, you can add and subtract in the numerator 3x2 and divide as indicated below. 7.2. Integration of simple fractions, As mentioned above, any improper rational fraction can be represented as the sum of some polynomial and a proper rational fraction (§7), and this representation is unique. Integrating a polynomial is not difficult, so consider the question of integrating a proper rational fraction. Since any proper rational fraction can be represented as a sum of simple fractions, its integration is reduced to the integration of simple fractions. Let us now consider the question of their integration. III. To find the integral of the simplest fraction of the third type, we isolate the complete square of the binomial from the square trinomial: Since the second term is equal to a2, where and then we make the substitution. Then, taking into account the linear properties of the integral, we find: Example 5. Find the integral 4 The integrand function is the simplest fraction of the third type, since the square trinomial x1 + Ax + 6 has no real roots (its discriminant is negative: , and the numerator contains a polynomial of the first degree. Therefore, we proceed as follows: 1) select a perfect square in the denominator 2) make a substitution (here 3) by * one integral To find the integral of the simplest fraction of the fourth type, we put, as above, . Then we get the Integral on the right side denoted by A and transform it as follows: Integral on the right side is integrated by parts, assuming from where or Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions Euler's first substitution Second Euler substitution Third substitution Euler We have obtained the so-called recurrent formula, which allows us to find the integral Jk for any k = 2, 3,. .. . Indeed, the integral J\ is tabular: Putting in the recurrence formula, we find Knowing and putting A = 3, we easily find Jj and so on. In the final result, substituting everywhere instead of t and a their expressions in terms of x and coefficients p and q, we obtain for the initial integral its expression in terms of x and the given numbers M, LG, p, q. Example 8. New integral “The integrand function is the simplest fraction of the fourth type, since the discriminant of a square trinomial is negative, i.e. This means that the denominator has no real roots, and the numerator is a polynomial of the 1st degree. 1) We select a complete square in the denominator 2) We make a substitution: The integral will take the form: Putting * = 2, a3 = 1 in the recurrence formula, we will have, and, therefore, the required integral is equal Returning to the variable x, we finally obtain 7.3. General case From the results of paragraphs. 1 and 2 of this section immediately follows an important theorem. Theorem! 4. The indefinite integral of any rational function always exists (on intervals in which the denominator of the fraction Q„(x) φ 0) and is expressed through a finite number of elementary functions, namely, it is an algebraic sum, the terms of which can only be multiplied , rational fractions, natural logarithms and arctangents. So, to find the indefinite integral of a fractional-rational function, one should proceed in the following way: 1) if the rational fraction is improper, then by dividing the numerator by the denominator, the whole part is isolated, i.e., this function is represented as the sum of a polynomial and a proper rational fraction; 2) then the denominator of the resulting proper fraction is decomposed into the product of linear and quadratic factors; 3) this proper fraction is decomposed into the sum of simple fractions; 4) using the linearity of the integral and the formulas of step 2, the integrals of each term are found separately. Example 7. Find the integral M Since the denominator is a polynomial of the third order, the integrand function is an improper fraction. We highlight the whole part in it: Therefore, we will have. The denominator of a proper fraction has phi different real roots: and therefore its decomposition into simple fractions has the form Hence we find. Giving the argument x values equal to the roots of the denominator, we find from this identity that: Consequently, the required integral will be equal to Example 8. Find the integral 4 The integrand is a proper fraction, the denominator of which has two different real roots: x - O multiplicity of 1 and x = 1 of multiplicity 3, Therefore, the expansion of the integrand into simple fractions has the form Bringing the right side of this equality to a common denominator and reducing both sides of the equality by this denominator, we obtain or. We equate the coefficients for the same powers of x on the left and right sides of this identity: From here we find. Substituting the found values of the coefficients into the expansion, we will have. Integrating, we find: Example 9. Find the integral 4 The denominator of the fraction has no real roots. Therefore, the expansion of the integrand into simple fractions has the form Hence or Equating the coefficients for the same powers of x on the left and right sides of this identity, we will have from where we find and, therefore, Remark. In the given example, the integrand function can be represented as a sum of simple fractions in a simpler way, namely, in the numerator of the fraction we select the binary that is in the denominator, and then we perform term-by-term division: §8. Integration of irrational functions A function of the form where Pm and £?„ are polynomials of degree type, respectively, in the variables uub2,... is called a rational function of ubu2j... For example, a polynomial of the second degree in two variables u\ and u2 has the form where - some real constants, and Example 1, The function is a rational function of the variables r and y, since it represents the ratio of a polynomial of the third degree and a polynomial of the fifth degree, but is not a yew function. In the case when the variables, in turn, are functions of the variable x: then the function ] is called a rational function of the functions of the Example. A function is a rational function of r and rvdikvlv Pryaivr 3. A function of the form is not a rational function of x and the radical y/r1 + 1, but it is a rational function of functions. As examples show, integrals of irrational functions are not always expressed through elementary functions. For example, integrals often encountered in applications are not expressed in terms of elementary functions; these integrals are called elliptic integrals of the first and second kind, respectively. Let us consider those cases when the integration of irrational functions can be reduced, with the help of some substitutions, to the integration of rational functions. 1. Let it be necessary to find the integral where R(x, y) is a rational function of its arguments x and y; m £ 2 - natural number; a, 6, c, d are real constants that satisfy the condition ad - bc ^ O (for ad - be = 0, the coefficients a and b are proportional to the coefficients c and d, and therefore the relationship does not depend on x; this means that in this case the integrand function will be a rational function of the variable x, the integration of which was discussed earlier). Let's make a change of variable in this integral, putting Hence we express the variable x through a new variable. We have x = - a rational function of t. Next we find or, after simplification, Therefore where A1 (t) is a rational function of *, since the rational funadia of a rational function, as well as the product of rational functions, are rational functions. We know how to integrate rational functions. Let Then the required integral be equal to At. IvYti integral 4 An integrand* function is a rational function of. Therefore, we set t = Then Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions Euler's first substitution Euler's second substitution Euler's third substitution Thus, we obtain Primar 5. Find the integral The common denominator of fractional exponents of x is equal to 12, so the integrand the function can be represented in the form 1 _ 1_ which shows that it is a rational function of: Taking this into account, let us put. Consequently, 2. Consider intephs of the form where the subintephal function is such that by replacing the radical \/ax2 + bx + c in it by y, we obtain a function R(x) y) - rational with respect to both arguments x and y. This integral is reduced to the integral of a rational function of another variable using Euler's substitutions. 8.1. Euler's first substitution Let the coefficient a > 0. Let us set or Hence we find x as a rational function of u, which means Thus, the indicated substitution expresses rationally in terms of *. Therefore, we will have a remark. The first Euler substitution can also be taken in the form Example 6. Let’s find the integral Therefore, we will have dx Euler’s substitution, show that Y 8.2. Euler's second substitution Let the trinomial ax2 + bx + c have different real roots R] and x2 (the coefficient can have any sign). In this case, we assume Since then we obtain Since x,dxn y/ax2 + be + c are expressed rationally in terms of t, then the original integral is reduced to the integral of a rational function, i.e. where Problem. Using Euler's first substitution, show that is a rational function of t. Example 7. Find the integral dx M function ] - x1 has different real roots. Therefore, we apply the second Euler substitution. From here we find. Substituting the found expressions into the Given?v*gyvl; we get 8.3. Third Euler substascom Let the coefficient c > 0. We make a change of variable by putting. Note that to reduce the integral to the integral of a rational function, the first and second Euler substitutions are sufficient. In fact, if the discriminant b2 -4ac > 0, then the roots of the quadratic trinomial ax + bx + c are real, and in this case the second Euler substitution is applicable. If, then the sign of the trinomial ax2 + bx + c coincides with the sign of the coefficient a, and since the trinomial must be positive, then a > 0. In this case, Euler’s first substitution is applicable. To find integrals of the type indicated above, it is not always advisable to use Euler’s substitutions, since for them it is possible to find other methods of integration that lead to the goal faster. Let's consider some of these integrals. 1. To find integrals of the form, isolate the right square from the square of the th trinomial: where After this, make a substitution and get where the coefficients a and P have different signs or they are both positive. For, and also for a > 0, the integral will be reduced to a logarithm, and if so, to the arcsine. At. Find imtegral 4 Sokak then. Assuming, we get Prmmar 9. Find. Assuming x -, we will have 2. The integral of the form is reduced to the integral y from step 1 as follows. Considering that the derivative ()" = 2, we highlight it in the numerator: 4 We identify the derivative of the radical expression in the numerator. Since (x, then we will have, taking into account the result of example 9, 3. Integrals of the form where P„(x) is a polynomial n -th degree, can be found by the method of indefinite coefficients, which consists of the following. Let us assume that the equality is Example 10. Mighty integral where Qn-i(s) is a polynomial of (n - 1) degree with indefinite coefficients: To find unknowns. coefficients | we differentiate both sides of (1): Then we reduce the right side of equality (2) to a common denominator equal to the denominator of the left side, i.e. y/ax2 + bx + c, reducing both sides of (2) by which we obtain the identity in both sides of which contain polynomials of degree n. Equating the coefficients for the same degrees of x in the left and right sides of (3), we obtain n + 1 equations, from which we find the required coefficients j4*(fc = 0,1,2,..., n ). Substituting their values into the right side of (1) and finding the integral + c we obtain the answer for this integral. Example 11. Find the integral Let's put Differentiating both suits of the equality, we will have Bringing the right side to a common denominator and reducing both sides by it, we get the identity or. Equating the coefficients at the same powers of x, we arrive at a system of equations from which we find = Then we find the integral on the right side of equality (4): Consequently, the required integral will be equal to
Complex integrals
This article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site.
It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles.
What integrals will be considered?
First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more.
Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way.
The third issue of the program will be integrals from complex fractions, which flew past the cash desk in previous articles.
Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.
(2) In the integrand function, we divide the numerator by the denominator term by term.
(3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign.
(4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since .
(5) We carry out a reverse replacement, expressing “te” from the direct replacement:
Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)
As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience.
In practice, of course, the square root is more common; here are three examples for solving it yourself:
Example 2
Find the indefinite integral
Example 3
Find the indefinite integral
Example 4
Find the indefinite integral
These examples are of the same type, so the complete solution at the end of the article will only be for Example 2; Examples 3-4 have the same answers. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose examples of the same type? Often found in their role. More often, perhaps, just something like 
.
But not always, when under the arctangent, sine, cosine, exponential and other functions there is a root of a linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be taken in an elementary way. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained.
By reducing the integral to itself
A witty and beautiful method. Let's take a look at the classics of the genre:
Example 5
Find the indefinite integral
Under the root is a quadratic binomial, and trying to integrate this example can give the teapot a headache for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how.
Let us denote the integral under consideration by a Latin letter and begin the solution: 
Let's integrate by parts: 
(1) Prepare the integrand function for term-by-term division.
(2) We divide the integrand function term by term. It may not be clear to everyone, but I’ll describe it in more detail:
(3) We use the linearity property of the indefinite integral.
(4) Take the last integral (“long” logarithm).
Now let's look at the very beginning of the solution: 
And at the end: 
What happened? As a result of our manipulations, the integral was reduced to itself!
Let's equate the beginning and the end: 
Move to the left side with a change of sign: 
And we move the two to the right side. As a result: 
The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here:
Note:
More strictly, the final stage of the solution looks like this:
Thus:
The constant can be redesignated by . Why can it be redesignated? Because he still accepts it any values, and in this sense there is no difference between constants and.
As a result:
A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself.
Example 6
Find the indefinite integral
Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example!
If under the square root there is a square trinomial, then the solution in any case comes down to two analyzed examples.
For example, consider the integral 
. All you need to do is first select a complete square:
.
Next, a linear replacement is carried out, which does “without any consequences”:
, resulting in the integral . Something familiar, right?
Or this example, with a quadratic binomial:
Select a complete square:
And, after linear replacement, we obtain the integral, which is also solved using the algorithm already discussed.
Let's look at two more typical examples of how to reduce an integral to itself:
– integral of the exponential multiplied by sine;
– integral of the exponential multiplied by the cosine.
In the listed integrals by parts you will have to integrate twice:
Example 7
Find the indefinite integral
The integrand is the exponential multiplied by the sine.
We integrate by parts twice and reduce the integral to itself: 
As a result of double integration by parts, the integral was reduced to itself. We equate the beginning and end of the solution:
We move it to the left side with a change of sign and express our integral: 
Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and in brackets place the sine and cosine in a “beautiful” order.
Now let's go back to the beginning of the example, or more precisely, to integration by parts: 
We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not necessarily. In fact, in the considered integral fundamentally doesn't matter, what do we mean by , we could have gone the other way: 
Why is this possible? Because the exponential turns into itself (both during differentiation and integration), sine and cosine mutually turn into each other (again, both during differentiation and integration).
That is, we can also denote a trigonometric function. But, in the example considered, this is less rational, since fractions will appear. If you wish, you can try to solve this example using the second method; the answers must match.
Example 8
Find the indefinite integral
This is an example for you to solve on your own. Before you decide, think about what is more advantageous in this case to designate as , an exponential or a trigonometric function? Full solution and answer at the end of the lesson.
And, of course, do not forget that most of the answers in this lesson are quite easy to check by differentiation!
The examples considered were not the most complex. In practice, integrals are more common where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will get confused in such an integral; I often get confused myself. The fact is that there is a high probability of fractions appearing in the solution, and it is very easy to lose something through carelessness. In addition, there is a high probability of an error in the signs; note that the exponent has a minus sign, and this introduces additional difficulty.
At the final stage, the result is often something like this:
Even at the end of the solution, you should be extremely careful and correctly understand the fractions: 
Integrating Complex Fractions
We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.
Continuing the theme of roots
Example 9
Find the indefinite integral
In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.
We decide: 
The replacement here is simple: 
Let's look at life after replacement: 
(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integrating Some Fractions, is being decided complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at straightening the result: under the root we again bring the terms to a common denominator and take them out from under the root.
Example 10
Find the indefinite integral 
This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same: 
The only thing you need to do additionally is to express the “x” from the replacement being carried out: 
Full solution and answer at the end of the lesson.
Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:
Example 11
Find the indefinite integral
Example 12
Find the indefinite integral
Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions.
Integral of an indecomposable polynomial of the 2nd degree to the power
(polynomial in denominator)
A more rare type of integral, but nevertheless encountered in practical examples.
Example 13
Find the indefinite integral
But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.
The solution starts with an artificial transformation: 
I think everyone already understands how to divide the numerator by the denominator term by term.
The resulting integral is taken in parts: 
For an integral of the form ( – natural number) we derive recurrent reduction formula:
, Where 
– integral of a degree lower.
Let us verify the validity of this formula for the solved integral.
In this case: , , we use the formula: 
As you can see, the answers are the same.
Example 14
Find the indefinite integral
This is an example for you to solve on your own. The sample solution uses the above formula twice in succession.
If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example:
What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand is expanded into a sum of fractions. But in my practice there is such an example never met, so I missed this case in the article Integrals of fractional-rational functions, I'll skip it now. If you still encounter such an integral, look at the textbook - everything is simple there. I don’t think it’s advisable to include material (even simple ones), the probability of encountering which tends to zero.
Integrating complex trigonometric functions
The adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high powers. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too.
In the above lesson we looked at universal trigonometric substitution for solving a certain type of integrals of trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided!
Let's consider another canonical example, the integral of one divided by sine:
Example 17
Find the indefinite integral
Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step:
(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: Divide in the denominator and multiply by .
(3) Using the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the differential sign.
(5) Take the integral.
A couple of simple examples for you to solve on your own:
Example 18
Find the indefinite integral
Note: The very first step should be to use the reduction formula 
and carefully carry out actions similar to the previous example.
Example 19
Find the indefinite integral
Well, this is a very simple example.
Complete solutions and answers at the end of the lesson.
I think now no one will have problems with integrals: 
etc.
What is the idea of the method? The idea is to use transformations and trigonometric formulas to organize only tangents and the tangent derivative into the integrand. That is, we are talking about replacing: 
. In Examples 17-19 we actually used this replacement, but the integrals were so simple that we got by with an equivalent action - subsuming the function under the differential sign.
Similar reasoning, as I already mentioned, can be carried out for the cotangent.
There is also a formal prerequisite for applying the above replacement:
The sum of the powers of cosine and sine is a negative integer EVEN number, For example:
for the integral – a negative integer EVEN number.
! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).
Let's look at a couple of more meaningful tasks based on this rule:
Example 20
Find the indefinite integral
The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative: 
(1) Let's transform the denominator.
(2) Using the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula 
.
(5) We bring the function under the differential sign.
(6) We carry out replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of getting confused.
Example 21
Find the indefinite integral
This is an example for you to solve on your own.
Hang in there, the championship rounds are about to begin =)
Often the integrand contains a “hodgepodge”:
Example 22
Find the indefinite integral 
This integral initially contains a tangent, which immediately leads to an already familiar thought:
I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above.
A couple of creative examples for your own solution:
Example 23
Find the indefinite integral 
Example 24
Find the indefinite integral 
Yes, in them, of course, you can lower the powers of sine and cosine, and use a universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson
An irrational function of a variable is a function that is formed from a variable and arbitrary constants using a finite number of operations of addition, subtraction, multiplication (raising to an integer power), division and taking roots. An irrational function differs from a rational one in that the irrational function contains operations for extracting roots.
There are three main types of irrational functions, the indefinite integrals of which are reduced to integrals of rational functions. These are integrals containing roots of arbitrary integer powers from a linear fractional function (the roots can be of different powers, but from the same linear fractional function); integrals of a differential binomial and integrals with the square root of a square trinomial.
Important note. Roots have multiple meanings!
When calculating integrals containing roots, expressions of the form are often encountered, where is some function of the integration variable. It should be borne in mind that. That is, at t > 0 , |t| = t. At t< 0 , |t| = - t . Therefore, when calculating such integrals, it is necessary to separately consider the cases t > 0 and t< 0 . This can be done by writing signs or wherever necessary. Assuming that the top sign refers to the case t > 0 , and the lower one - to the case t< 0 . With further transformation, these signs, as a rule, cancel each other.
A second approach is also possible, in which the integrand and the result of integration can be considered as complex functions of complex variables. Then you don’t have to pay attention to the signs in radical expressions. This approach is applicable if the integrand is analytic, that is, a differentiable function of a complex variable. In this case, both the integrand and its integral are multivalued functions. Therefore, after integration, when substituting numerical values, it is necessary to select a single-valued branch (Riemann surface) of the integrand, and for it select the corresponding branch of the integration result.
Fractional linear irrationality
These are integrals with roots from the same fractional linear function:
,
where R is a rational function, are rational numbers, m 1, n 1, ..., m s, n s are integers, α, β, γ, δ are real numbers.
Such integrals are reduced to the integral of a rational function by substitution:
, where n is the common denominator of the numbers r 1, ..., r s.
The roots may not necessarily come from a linear fractional function, but also from a linear one (γ = 0 , δ = 1), or on the integration variable x (α = 1, β = 0, γ = 0, δ = 1).
Here are examples of such integrals:
,
.
Integrals from differential binomials
Integrals from differential binomials have the form:
,
where m, n, p are rational numbers, a, b are real numbers.
Such integrals reduce to integrals of rational functions in three cases.
1) If p is an integer. Substitution x = t N, where N is the common denominator of the fractions m and n.
2) If - an integer. Substitution a x n + b = t M, where M is the denominator of the number p.
3) If - an integer. Substitution a + b x - n = t M, where M is the denominator of the number p.
In other cases, such integrals are not expressed through elementary functions.
Sometimes such integrals can be simplified using reduction formulas:
;
.
Integrals containing the square root of a square trinomial
Such integrals have the form:
,
where R is a rational function. For each such integral there are several methods for solving it.
1)
Using transformations lead to simpler integrals.
2)
Apply trigonometric or hyperbolic substitutions.
3)
Apply Euler substitutions.
Let's look at these methods in more detail.
1) Transformation of the integrand function
Applying the formula and performing algebraic transformations, we reduce the integrand function to the form:
,
where φ(x), ω(x) are rational functions.
Type I
Integral of the form:
,
where P n (x) is a polynomial of degree n.
Such integrals are found by the method of indefinite coefficients using the identity:
.
Differentiating this equation and equating the left and right sides, we find the coefficients A i.
Type II
Integral of the form:
,
where P m (x) is a polynomial of degree m.
Substitution t = (x - α) -1 this integral is reduced to the previous type. If m ≥ n, then the fraction should have an integer part.
III type
Here we do the substitution:
.
After which the integral will take the form:
.
Next, the constants α, β must be chosen such that the coefficients of t in the denominator become zero:
B = 0, B 1 = 0.
Then the integral decomposes into the sum of integrals of two types:
,
,
which are integrated by substitutions:
u 2 = A 1 t 2 + C 1,
v 2 = A 1 + C 1 t -2 .
2) Trigonometric and hyperbolic substitutions
For integrals of the form , a > 0
,
we have three main substitutions:
;
;
;
For integrals, a > 0
,
we have the following substitutions:
;
;
;
And finally, for the integrals, a > 0
,
the substitutions are as follows:
;
;
;
3) Euler substitutions
Also, integrals can be reduced to integrals of rational functions of one of three Euler substitutions:
, for a > 0;
, for c > 0 ;
, where x 1 is the root of the equation a x 2 + b x + c = 0. If this equation has real roots.
Elliptic integrals
In conclusion, consider integrals of the form:
,
where R is a rational function, . Such integrals are called elliptic. In general, they are not expressed through elementary functions. However, there are cases when there are relationships between the coefficients A, B, C, D, E, in which such integrals are expressed through elementary functions.
Below is an example related to reflexive polynomials. The calculation of such integrals is performed using substitutions:
.
Example
Calculate the integral:
.
Solution
Let's make a substitution.
.
Here at x > 0
(u> 0
) take the upper sign ′+ ′. At x< 0
(u< 0
) - lower ′- ′.
.
Answer
Used literature:
N.M. Gunter, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.
There is no universal way to solve irrational equations, since their class differs in quantity. The article will highlight characteristic types of equations with substitution using the integration method.
To use the direct integration method, it is necessary to calculate indefinite integrals of the type ∫ k x + b p d x , where p is a rational fraction, k and b are real coefficients.
Example 1
Find and calculate the antiderivatives of the function y = 1 3 x - 1 3 .
Solution
According to the integration rule, it is necessary to apply the formula ∫ f (k x + b) d x = 1 k F (k x + b) + C, and the table of antiderivatives indicates that there is a ready-made solution to this function. We get that
∫ d x 3 x - 1 3 = ∫ (3 x - 1) - 1 3 d x = 1 3 1 - 1 3 + 1 (3 x - 1) - 1 3 + 1 + C = = 1 2 (3 x - 1) 2 3 + C
Answer:∫ d x 3 x - 1 3 = 1 2 (3 x - 1) 2 3 + C .
There are cases when it is possible to use the method of subsuming a differential sign. This is solved by the principle of finding indefinite integrals of the form ∫ f " (x) · (f (x)) p d x , when the value of p is considered a rational fraction.
Example 2
Find the indefinite integral ∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x .
Solution
Note that d x 3 + 5 x - 7 = x 3 + 5 x - 7 "d x = (3 x 2 + 5) d x. Then it is necessary to subsume the differential sign using tables of antiderivatives. We obtain that
∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x = ∫ (x 3 + 5 x - 7) - 7 6 (3 x 2 + 5) d x = = ∫ (x 3 + 5 x - 7 ) - 7 6 d (x 3 + 5 x - 7) = x 3 + 5 x - 7 = z = = ∫ z - 7 6 d z = 1 - 7 6 + 1 z - 7 6 + 1 + C = - 6 z - 1 6 + C = z = x 3 + 5 x - 7 = - 6 (x 3 + 5 x - 7) 6 + C
Answer:∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x = - 6 (x 3 + 5 x - 7) 6 + C .
Solving indefinite integrals involves a formula of the form ∫ d x x 2 + p x + q, where p and q are real coefficients. Then you need to select a complete square from under the root. We get that
x 2 + p x + q = x 2 + p x + p 2 2 - p 2 2 + q = x + p 2 2 + 4 q - p 2 4
Applying the formula located in the table of indefinite integrals, we obtain:
∫ d x x 2 ± α = ln x + x 2 ± α + C
Then the integral is calculated:
∫ d x x 2 + p x + q = ∫ d x x + p 2 2 + 4 q - p 2 4 = = ln x + p 2 + x + p 2 2 + 4 q - p 2 4 + C = = ln x + p 2 + x 2 + p x + q + C
Example 3
Find the indefinite integral of the form ∫ d x 2 x 2 + 3 x - 1 .
Solution
To calculate, you need to take out the number 2 and place it in front of the radical:
∫ d x 2 x 2 + 3 x - 1 = ∫ d x 2 x 2 + 3 2 x - 1 2 = 1 2 ∫ d x x 2 + 3 2 x - 1 2
Select a complete square in radical expression. We get that
x 2 + 3 2 x - 1 2 = x 2 + 3 2 x + 3 4 2 - 3 4 2 - 1 2 = x + 3 4 2 - 17 16
Then we obtain an indefinite integral of the form 1 2 ∫ d x x 2 + 3 2 x - 1 2 = 1 2 ∫ d x x + 3 4 2 - 17 16 = = 1 2 ln x + 3 4 + x 2 + 3 2 x - 1 2 + C
Answer: d x x 2 + 3 x - 1 = 1 2 ln x + 3 4 + x 2 + 3 2 x - 1 2 + C
Integration of irrational functions is carried out in a similar way. Applicable for functions of the form y = 1 - x 2 + p x + q.
Example 4
Find the indefinite integral ∫ d x - x 2 + 4 x + 5 .
Solution
First you need to derive the square of the denominator of the expression from under the root.
∫ d x - x 2 + 4 x + 5 = ∫ d x - x 2 - 4 x - 5 = = ∫ d x - x 2 - 4 x + 4 - 4 - 5 = ∫ d x - x - 2 2 - 9 = ∫ d x - (x - 2) 2 + 9
The table integral has the form ∫ d x a 2 - x 2 = a r c sin x a + C, then we obtain that ∫ d x - x 2 + 4 x + 5 = ∫ d x - (x - 2) 2 + 9 = a r c sin x - 2 3 +C
Answer:∫ d x - x 2 + 4 x + 5 = a r c sin x - 2 3 + C .
The process of finding antiderivative irrational functions of the form y = M x + N x 2 + p x + q, where the existing M, N, p, q are real coefficients, and are similar to the integration of simple fractions of the third type. This transformation has several stages:
summing the differential under the root, isolating the complete square of the expression under the root, using tabular formulas.
Example 5
Find the antiderivatives of the function y = x + 2 x 2 - 3 x + 1.
Solution
From the condition we have that d (x 2 - 3 x + 1) = (2 x - 3) d x and x + 2 = 1 2 (2 x - 3) + 7 2, then (x + 2) d x = 1 2 (2 x - 3) + 7 2 d x = 1 2 d (x 2 - 3 x + 1) + 7 2 d x .
Let's calculate the integral: ∫ x + 2 x 2 - 3 x + 1 d x = 1 2 ∫ d (x 2 - 3 x + 1) x 2 - 3 x + 1 + 7 2 ∫ d x x 2 - 3 x + 1 = = 1 2 ∫ (x 2 - 3 x + 1) - 1 2 d (x 2 - 3 x + 1) + 7 2 ∫ d x x - 3 2 2 - 5 4 = = 1 2 1 - 1 2 + 1 x 2 - 3 x + 1 - 1 2 + 1 + 7 2 ln x - 3 2 + x - 3 2 - 5 4 + C = = x 2 - 3 x + 1 + 7 2 ln x - 3 2 + x 2 - 3 x + 1 + C
Answer:∫ x + 2 x 2 - 3 x + 1 d x = x 2 - 3 x + 1 + 7 2 ln x - 3 2 + x 2 - 3 x + 1 + C .
The search for indefinite integrals of the function ∫ x m (a + b x n) p d x is carried out using the substitution method.
To solve it is necessary to introduce new variables:
- When p is an integer, then x = z N is considered, and N is the common denominator for m, n.
- When m + 1 n is an integer, then a + b x n = z N, and N is the denominator of p.
- When m + 1 n + p is an integer, then the variable a x - n + b = z N is required, and N is the denominator of the number p.
Find the definite integral ∫ 1 x 2 x - 9 d x .
Solution
We get that ∫ 1 x 2 x - 9 d x = ∫ x - 1 · (- 9 + 2 x 1) - 1 2 d x . It follows that m = - 1, n = 1, p = - 1 2, then m + 1 n = - 1 + 1 1 = 0 is an integer. You can introduce a new variable of the form - 9 + 2 x = z 2. It is necessary to express x in terms of z. As output we get that
9 + 2 x = z 2 ⇒ x = z 2 + 9 2 ⇒ d x = z 2 + 9 2 " d z = z d z - 9 + 2 x = z
It is necessary to make a substitution into the given integral. We have that
∫ d x x 2 x - 9 = ∫ z d z z 2 + 9 2 z = 2 ∫ d z z 2 + 9 = = 2 3 a r c t g z 3 + C = 2 3 a r c c t g 2 x - 9 3 + C
Answer:∫ d x x 2 x - 9 = 2 3 a r c c t g 2 x - 9 3 + C .
To simplify the solution of irrational equations, basic integration methods are used.
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