Integrals of fractionally irrational functions. Integration - MT1205: Mathematical Analysis for Economists - Business Informatics

We continue to consider integrals of fractions and roots. 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.

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.

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, we rearranged the terms into convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.

(4) The resulting integral, as you remember, is solved 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 independent decision. Here a constant is added to the lone “X”, and the replacement is almost the same:

The only thing that is needed is to additionally 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 of which was discussed in class Integrals from irrational functions .

Integral of a polynomial of the 2nd degree indecomposable in the denominator to the power

A more rare type of integral, but nevertheless encountered in practical examples.

Example 13

Find the indefinite integral

The denominator of the integrand contains a quadratic binomial that cannot be factorized. We emphasize that non-factorizability is an essential feature. If the polynomial is factorized, then everything is much clearer, for example:

Let's go back to the example with lucky number 13. This integral is also one of those that can be quite painful 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

Where ( k≥ 2) – natural number, derived recurrent reduction formula:

; is an integral of a degree lower by 1.

What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand function is expanded into a sum of fractions. If you encounter such an integral, look at the textbook - everything is simple there.

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 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 the 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.

Example 6

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.

If you notice an error in the text, please highlight it and press Ctrl+Enter

The class of irrational functions is very wide, so there simply cannot be a universal way to integrate them. In this article we will try to identify the most characteristic types of irrational integrand functions and associate the integration method with them.

There are cases when it is appropriate to use the method of subscribing to the differential sign. For example, when finding indefinite integrals of the form, where p– rational fraction.

Example.

Find the indefinite integral .

Solution.

It is not difficult to notice that . Therefore, we put it under the differential sign and use the table of antiderivatives:

Answer:

13. Fractional linear substitution

Integrals of the type where a, b, c, d are real numbers, a, b,..., d, g are natural numbers, are reduced to integrals of a rational function by substitution, where K is the least common multiple of the denominators of the fractions

Indeed, from the substitution it follows that

i.e. x and dx are expressed through rational functions of t. Moreover, each degree of the fraction is expressed through a rational function of t.

Example 33.4. Find the integral

Solution: The least common multiple of the denominators of the fractions 2/3 and 1/2 is 6.

Therefore, we put x+2=t 6, x=t 6 -2, dx=6t 5 dt, Therefore,

Example 33.5. Specify the substitution for finding integrals:

Solution: For I 1 substitution x=t 2, for I 2 substitution

14. Trigonometric substitution

Integrals of type are reduced to integrals of functions that rationally depend on trigonometric functions using the following trigonometric substitutions: x = a sint for the first integral; x=a tgt for the second integral; for the third integral.

Example 33.6. Find the integral

Solution: Let's put x=2 sin t, dx=2 cos tdt, t=arcsin x/2. Then

Here the integrand is a rational function with respect to x and By selecting a complete square under the radical and making a substitution, integrals of the indicated type are reduced to integrals of the type already considered, i.e., to integrals of the type These integrals can be calculated using appropriate trigonometric substitutions.

Example 33.7. Find the integral

Solution: Since x 2 +2x-4=(x+1) 2 -5, then x+1=t, x=t-1, dx=dt. That's why Let's put

Note: Integral type It is expedient to find using the substitution x=1/t.

15. Definite integral

Let a function be defined on a segment and have an antiderivative on it. The difference is called definite integral functions along the segment and denote. So,

The difference is written in the form, then . Numbers are called limits of integration .

For example, one of the antiderivatives for a function. That's why

16 . If c is a constant number and the function ƒ(x) is integrable on , then

that is, the constant factor c can be taken out of the sign of the definite integral.

▼Let’s compose the integral sum for the function with ƒ(x). We have:

Then it follows that the function c ƒ(x) is integrable on [a; b] and formula (38.1) is valid.▲

2. If the functions ƒ 1 (x) and ƒ 2 (x) are integrable on [a;b], then integrable on [a; b] their sum u

that is, the integral of the sum is equal to the sum of the integrals.

▼
▲

Property 2 applies to the sum of any finite number of terms.

This property can be accepted by definition. This property is also confirmed by the Newton-Leibniz formula.

4. If the function ƒ(x) is integrable on [a; b] and a< с < b, то

that is, the integral over the entire segment is equal to the sum of the integrals over the parts of this segment. This property is called the additivity of a definite integral (or the additivity property).

When dividing the segment [a;b] into parts, we include point c in the number of division points (this can be done due to the independence of the limit of the integral sum from the method of dividing the segment [a;b] into parts). If c = x m, then the integral sum can be divided into two sums:

Each of the written sums is integral, respectively, for the segments [a; b], [a; s] and [s; b]. Passing to the limit in the last equality as n → ∞ (λ → 0), we obtain equality (38.3).

Property 4 is valid for any location of points a, b, c (we assume that the function ƒ (x) is integrable on the larger of the resulting segments).

So, for example, if a< b < с, то

(properties 4 and 3 are used).

5. “Theorem on mean values.” If the function ƒ(x) is continuous on the interval [a; b], then there is a tonka with є [a; b] such that

▼By the Newton-Leibniz formula we have

where F"(x) = ƒ(x). Applying the Lagrange theorem (the theorem on the finite increment of a function) to the difference F(b)-F(a), we obtain

F(b)-F(a) = F"(c) (b-a) = ƒ(c) (b-a).▲

Property 5 (“the mean value theorem”) for ƒ (x) ≥ 0 has a simple geometric meaning: the value of the definite integral is equal, for some c є (a; b), to the area of a rectangle with height ƒ (c) and base b-a ( see fig. 170). Number

is called the average value of the function ƒ(x) on the interval [a; b].

6. If the function ƒ (x) maintains its sign on the segment [a; b], where a< b, то интегралимеет тот же знак, что и функция. Так, если ƒ(х)≥0 на отрезке [а; b], то

▼By the “mean value theorem” (property 5)

where c є [a; b]. And since ƒ(x) ≥ 0 for all x О [a; b], then

ƒ(с)≥0, b-а>0.

Therefore ƒ(с) (b-а) ≥ 0, i.e. ▲

7. Inequality between continuous functions on the interval [a; b], (a

▼Since ƒ 2 (x)-ƒ 1 (x)≥0, then when a< b, согласно свойству 6, имеем

Or, according to property 2,

Note that it is impossible to differentiate inequalities.

8. Estimation of the integral. If m and M are, respectively, the smallest and largest values of the function y = ƒ (x) on the segment [a; b], (a< b), то

▼Since for any x є [a;b] we have m≤ƒ(x)≤М, then, according to property 7, we have

Applying Property 5 to the extreme integrals, we obtain

▲

If ƒ(x)≥0, then property 8 is illustrated geometrically: the area of a curvilinear trapezoid is enclosed between the areas of rectangles whose base is , and whose heights are m and M (see Fig. 171).

9. The modulus of a definite integral does not exceed the integral of the modulus of the integrand:

▼Applying property 7 to the obvious inequalities -|ƒ(x)|≤ƒ(x)≤|ƒ(x)|, we obtain

It follows that

▲

10. The derivative of a definite integral with respect to a variable upper limit is equal to the integrand in which the integration variable is replaced by this limit, i.e.

Calculating the area of a figure is one of the most difficult problems in area theory. In the school geometry course, we learned to find the areas of basic geometric shapes, for example, a circle, triangle, rhombus, etc. However, much more often you have to deal with calculating the areas of more complex figures. When solving such problems, one has to resort to integral calculus.

In this article we will consider the problem of calculating the area of a curvilinear trapezoid, and we will approach it in a geometric sense. This will allow us to find out the direct connection between the definite integral and the area of a curvilinear trapezoid.

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 with 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 the system 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 and has the form Given right side to a common denominator, we will 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 equal zero 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 decompositions 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, given linear properties integral, we find: Example 5. Find the integral 4 The integrand 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 the perfect square in the denominator 2) make a substitution (here 3) for * 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 can easily find Jj and so on. In the final result, substituting everywhere instead of t and a their expressions in terms of x and the coefficients p and q, we obtain for the original 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 in the recurrence formula * = 2, a3 = 1. 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 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: Therefore, 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 example given, the integrand can be represented as a sum of simple fractions of more than in a simple way , namely, in the numerator of the fraction we select the binomial 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 z and y, since it represents both the ratio of a polynomial of the third degree and a polynomial of the fifth degree and 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 First Euler substitution Second Euler substitution Third Euler substitution Thus, we obtain Primar 5. Find the integral Common denominator fractional indicators powers of x is 12, so the integrand can be represented as 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 reduces to the integral of a rational function of another variable substitutions Euler. 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 look at some of these integrals. 1. To find integrals of the form, select a perfect 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

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 comprehensive functions from 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 in the denominator the coefficients for t 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 view 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. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.