Integration of an irrational function of the form r x. Integration of irrational expressions

Under irrational understand an expression in which the independent variable %% x %% or the polynomial %% P_n (x) %% of degree %% n \\ in \\ mathbb (N) %% are included under the sign radical (from latin radix - root), i.e. are raised to a fractional power. Some classes of integrands that are irrational with respect to %% x %% can be reduced by changing the variable to rational expressions with respect to the new variable.

The notion of a rational function of one variable can be extended to several arguments. If over each argument %% u, v, \\ dotsc, w %% when calculating the value of a function, only arithmetic operations and raising to an integer power are provided, then we speak of a rational function of these arguments, which is usually denoted %% R (u, v, \\ The arguments of such a function can themselves be functions of the independent variable %% x %%, including radicals like %% \\ sqrt [n] (x), n \\ in \\ mathbb (N) %%. For example, the rational function $$ R (u, v, w) \u003d \\ frac (u + v ^ 2) (w) $$ for %% u \u003d x, v \u003d \\ sqrt (x) %% and %% w \u003d \\ sqrt (x ^ 2 + 1) %% is a rational function $$ R \\ left (x, \\ sqrt (x), \\ sqrt (x ^ 2 + 1) \\ right) \u003d \\ frac (x + \\ sqrt (x ^ 2)) (\\ sqrt (x ^ 2 + 1)) \u003d f (x) $$ from %% x %% and radicals %% \\ sqrt (x) %% and %% \\ sqrt (x ^ 2 + 1 ) %%, while the function %% f (x) %% will be an irrational (algebraic) function of one independent variable %% x %%.

Consider integrals of the form %% \\ int R (x, \\ sqrt [n] (x)) \\ mathrm (d) x %%. Such integrals are rationalized by changing the variable %% t \u003d \\ sqrt [n] (x) %%, then %% x \u003d t ^ n, \\ mathrm (d) x \u003d nt ^ (n-1) %%.

Example 1

Find %% \\ displaystyle \\ int \\ frac (\\ mathrm (d) x) (\\ sqrt (x) + \\ sqrt (x)) %%.

The integrand of the required argument is written as a function of radicals of degree %% 2 %% and %% 3 %%. Since the least common multiple of the numbers %% 2 %% and %% 3 %% is equal to %% 6 %%, this integral is an integral like %% \\ int R (x, \\ sqrt (x)) \\ mathrm (d) x %% and can be rationalized by replacing %% \\ sqrt (x) \u003d t %%. Then %% x \u003d t ^ 6, \\ mathrm (d) x \u003d 6t \\ mathrm (d) t, \\ sqrt (x) \u003d t ^ 3, \\ sqrt (x) \u003d t ^ 2 %%. Therefore, $$ \\ int \\ frac (\\ mathrm (d) x) (\\ sqrt (x) + \\ sqrt (x)) \u003d \\ int \\ frac (6t ^ 5 \\ mathrm (d) t) (t ^ 3 + t ^ 2) \u003d 6 \\ int \\ frac (t ^ 3) (t + 1) \\ mathrm (d) t. $$ Take %% t + 1 \u003d z, \\ mathrm (d) t \u003d \\ mathrm (d) z, z \u003d t + 1 \u003d \\ sqrt (x) + 1 %% and $$ \\ begin (array) (ll ) \\ int \\ frac (\\ mathrm (d) x) (\\ sqrt (x) + \\ sqrt (x)) & \u003d 6 \\ int \\ frac ((z-1) ^ 3) (z) \\ mathrm (d) t \u003d \\\\ & \u003d 6 \\ int z ^ 2 dz -18 \\ int z \\ mathrm (d) z + 18 \\ int \\ mathrm (d) z -6 \\ int \\ frac (\\ mathrm (d) z) (z ) \u003d \\\\ & \u003d 2z ^ 3 - 9 z ^ 2 + 18z -6 \\ ln | z | + C \u003d \\\\ & \u003d 2 \\ left (\\ sqrt (x) + 1 \\ right) ^ 3 - 9 \\ left (\\ sqrt (x) + 1 \\ right) ^ 2 + \\\\ & + ~ 18 \\ left ( \\ sqrt (x) + 1 \\ right) - 6 \\ ln \\ left | \\ sqrt (x) + 1 \\ right | + C \\ end (array) $$

Integrals of the form %% \\ int R (x, \\ sqrt [n] (x)) \\ mathrm (d) x %% are a special case of linear fractional irrationalities, i.e. integrals of the form %% \\ displaystyle \\ int R \\ left (x, \\ sqrt [n] (\\ dfrac (ax + b) (cd + d)) \\ right) \\ mathrm (d) x %%, where %% ad - bc \\ neq 0 %%, which can be rationalized by changing the variable %% t \u003d \\ sqrt [n] (\\ dfrac (ax + b) (cd + d)) %%, then %% x \u003d \\ dfrac (dt ^ n - b) (a - ct ^ n) %%. Then $$ \\ mathrm (d) x \u003d \\ frac (n t ^ (n-1) (ad - bc)) (\\ left (a - ct ^ n \\ right) ^ 2) \\ mathrm (d) t. $$

Example 2

Find %% \\ displaystyle \\ int \\ sqrt (\\ dfrac (1 -x) (1 + x)) \\ dfrac (\\ mathrm (d) x) (x + 1) %%.

Let's take %% t \u003d \\ sqrt (\\ dfrac (1 -x) (1 + x)) %%, then %% x \u003d \\ dfrac (1 - t ^ 2) (1 + t ^ 2) %%, $$ \\ begin (array) (l) \\ mathrm (d) x \u003d - \\ frac (4t \\ mathrm (d) t) (\\ left (1 + t ^ 2 \\ right) ^ 2), \\\\ 1 + x \u003d \\ \\ end (array) $$ Therefore, $$ \\ begin (array) (l) \\ int \\ sqrt (\\ dfrac (1 -x) (1 + x)) \\ frac (\\ mathrm (d) x) (x + 1) \u003d \\\\ \u003d \\ frac (t (1 + t ^ 2)) (2) \\ left (- \\ frac (4t \\ mathrm (d) t) (\\ left (1 + t ^ 2 \\ right) ^ 2 ) \\ right) \u003d \\\\ \u003d -2 \\ int \\ frac (t ^ 2 \\ mathrm (d) t) (1 + t ^ 2) \u003d \\\\ \u003d -2 \\ int \\ mathrm (d) t + 2 \\ int \\ frac (\\ mathrm (d) t) (1 + t ^ 2) \u003d \\\\ \u003d -2t + \\ text (arctg) ~ t + C \u003d \\\\ \u003d -2 \\ sqrt (\\ dfrac (1 -x) ( 1 + x)) + \\ text (arctg) ~ \\ sqrt (\\ dfrac (1 -x) (1 + x)) + C. \\ end (array) $$

Consider integrals of the form %% \\ int R \\ left (x, \\ sqrt (ax ^ 2 + bx + c) \\ right) \\ mathrm (d) x %%. In the simplest cases, such integrals are reduced to tabular integrals if, after isolating a complete square, we make a change of variables.

Example 3

Find the integral %% \\ displaystyle \\ int \\ dfrac (\\ mathrm (d) x) (\\ sqrt (x ^ 2 + 4x + 5)) %%.

Considering that %% x ^ 2 + 4x + 5 \u003d (x + 2) ^ 2 + 1 %%, we assume %% t \u003d x + 2, \\ mathrm (d) x \u003d \\ mathrm (d) t %%, then $$ \\ begin (array) (ll) \\ int \\ frac (\\ mathrm (d) x) (\\ sqrt (x ^ 2 + 4x + 5)) & \u003d \\ int \\ frac (\\ mathrm (d) t) (\\ sqrt (t ^ 2 + 1)) \u003d \\\\ & \u003d \\ ln \\ left | t + \\ sqrt (t ^ 2 + 1) \\ right | + C \u003d \\\\ & \u003d \\ ln \\ left | x + 2 + \\ sqrt (x ^ 2 + 4x + 5) \\ right | + C. \\ end (array) $$

In more difficult cases to find integrals of the form %% \\ int R \\ left (x, \\ sqrt (ax ^ 2 + bx + c) \\ right) \\ mathrm (d) x %% use

This section will consider the method of integrating rational functions. 7.1. Brief information on rational functions The simplest rational function is a polynomial of the ti-th degree, i.e. a function of the form where are real constants, and a0 Φ 0. The polynomial Qn (x), for which the coefficient a0 \u003d 1 "is called reduced. A real number b is called a root of the polynomial Qn (z) if Q „(b) \u003d 0. It is known that every polynomial Qn (x) with real coefficients unique way decomposes into real factors of the form where p, q are real coefficients, and the quadratic factors do not have real roots and, therefore, are indecomposable into real linear factors. Combining the same factors (if any) and assuming, for simplicity, the polynomial Qn (x) 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, the sum of all exponents a, / 3, ..., A, added with the doubled sum of all exponents ui, ..., q, is equal to n: The root a of the polynomial is called simple or single if a \u003d 1, and multiple if a\u003e 1; the number a is called the multiplicity of the root a. The same is true for the 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 correct 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 incorrect and in this case, dividing the numerator by the denominator according to the rule of dividing polynomials, it can be represented in the form where are some polynomials, and ^^ is a regular rational fraction. Example 1. A rational fraction is an irregular fraction. Dividing by "corner", we will have Consequently. Here. and a correct fraction. Definition. The simplest (or elementary) fractions are rational fractions of the following four types: where - 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 that -2 _2 is its discriminant. The following theorem is proved in algebra. Theorem 3. A regular rational fraction with real coefficients, the denominator of which Qn (x) is decomposed in a unique way into a sum of elementary fractions according to the rule Integration of rational functions Brief information about rational functions Integration of elementary 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 reduced 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 desired constants are found. ... This method of finding unknown constants is called the undefined coefficient method. Sometimes it is more convenient to apply another method for finding unknown constants, which consists in the fact that after equating the numerators, an identity with respect to x is obtained, in which some values \u200b\u200bare assigned to the argument x, for example, the values \u200b\u200bof the roots, as a result of which equations for finding the constants are obtained. It is especially convenient if the denominator Q „(x) has only real simple roots. Example 2. Expand a rational fraction into simple fractions This fraction is regular. We decompose the denominator into factors of ate: Since the roots of the denominator are real and different, then on the basis of formula (1) the decomposition of the fraction into the simplest will have the form of The unknowns to the coefficient A. 2?, C are found in two ways. The first way. Equating the coefficients at the same powers of x, i.e. at (free term), and the left and right sides of the identity, we obtain linear system equations for finding unknown coefficients A, B, C: This system has a unique solution C Second way. Tek as the roots of the denominator are torn stv in i 0, we get 2 \u003d 2A, whence A * 1; r i 1, we get -1 * -B, whence 5 * 1; x i 2, we get 2 \u003d 2C. whence С »1, and the required decomposition has the form 3. Reinstate not the simplest fractions, rational fraction 4 We decompose the polynomial, standing at anaive, into factors:. The denominator has two different double roots: x \\ \u003d 0 of multiplicity 3. Therefore, the expansion of this fraction is not simpler is of the form Bringing the right-hand side to a common denominator, we find or the First way. Equating the coefficients at 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, setting x \u003d 0, we obtain 1 a A2, or A2 \u003d 1; field * gay x \u003d -1, we get -3 i B), or Bj i -3. When substituting the found values \u200b\u200bof the coefficients A \\ and B) a, the identity will take the form or Assuming x \u003d 0, and then x \u003d -I. we find that \u003d 0, B2 \u003d 0 and. therefore, B \\ \u003d 0. Thus, we again obtain Example 4. Expand the rational fraction into simple fractions 4 The denominator of the fraction has no real roots, since the function x2 + 1 does not vanish for any real values \u200b\u200bof x. Therefore, the expansion into elementary fractions should have the form From this we obtain or. Equating the coefficients at Sshinak powers of x in the left and right sides of the last equality, we will have from 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 expansion of the fraction in example 3, you can add and subtract in the numerator Zx2 and perform division, as indicated below. 7.2. Integration of elementary fractions. As mentioned above, any irregular rational fraction can be represented as the sum of a certain polynomial and a regular rational fraction (§7), and this representation is unique. Integration of a polynomial is not difficult, therefore, we will consider the question of integrating a regular rational fraction. Since any regular rational fraction can be represented as a sum of the simplest fractions, its integration is reduced to the integration of the simplest fractions. Let us now consider the question of their integration. III. To find the integral of the simplest fraction of the third type, let us select the complete square of the binomial from the quadratic trinomial: Since the second term we set it equal to a2, where and then we will make a substitution. Then, taking into account the linear properties of the 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) we select a complete square in the denominator 2) we 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-hand side, denote it by A and transform it as follows: We integrate the integral on the right-hand side by parts, setting whence or 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 substitution Euler We have obtained the so-called recurrent formula, which allows finding the integral Jk for any k \u003d 2, 3, .... Indeed, the integral J \\ is tabular: Assuming in the recurrent formula, we find Knowing and setting A \u003d 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 initial integral its expression in terms of x and the given numbers M, AH, p, q. Example 8. Neyti integral “The integrable function is the simplest fraction of the fourth type, since the discriminant of the square trinomial is negative, that is, therefore, the denominator has no real roots, and the numerator is a polynomial of the 1st degree. 1) Allocate a full square in the denominator 2) We make the substitution: The integral takes the form: Assuming in the recurrent formula * \u003d 2, a3 \u003d 1. we will have, and, therefore, the required integral is equal. Returning to the variable x, we finally get 7.3. General case From the results of Sec. 1 and 2 of this section an important theorem immediately follows. Theorem! 4. An indefinite integral of any rational function always exists (on intervals in which the denominator of the fraction Qn (x) Φ 0) and is expressed in terms of a finite number of elementary functions, namely, it is an algebraic sum, whose members can only be multiplied , rational fractions, natural logarithms and arctangents. So, to find the indefinite integral of a fractional-rational function, one should act in the following way: 1) if the rational fraction is incorrect, then the whole part is allocated by dividing the numerator by the denominator, i.e. that is, this function is represented as a sum of a polynomial and a regular rational fraction; 2) then the denominator of the obtained correct fraction is decomposed into the product of linear and quadratic factors; 3) this regular fraction is decomposed into the sum of the simplest fractions; 4) using the linearity of the integral and the formulas of item 2, the integrals of each term are found separately. Example 7. Find the integral М Since the denominator is a polynomial of the third step, the integrand is an irregular fraction. We single out the whole part in it: Therefore, we will have. The denominator of a regular fraction has phi of different real roots: and therefore, its decomposition into simple fractions has the form From this we find. Giving the argument x values \u200b\u200bequal 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 regular fraction, the denominator of which has two different real roots: x - О of multiplicity 1 and х \u003d 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 canceling both sides of the equality by this denominator, we get or. We equate the coefficients at the same powers of x in the left and right sides of this identity: From here we find. Substituting the found values \u200b\u200bof the coefficients in 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 into the simplest fractions of the integrand has the form From here or By equating the coefficients at the same powers of x in the left and right sides of this identity, we will have whence we find and, therefore, Remark. In the given example, the integrand can be represented as a sum of elementary fractions in a simpler way, namely, in the numerator of the fraction we select the binomial in the denominator, and then we perform term-by-term division: §8. Integration of irrational functions A function of the form where Pm and U2 are polynomials of degree type, respectively, in variables ub2, ... is called a different function of ubu2j ... For example, a polynomial of second degree in two variables u1 and u2 has the form where are some real constants, and Example 1, The function is a rational function of the variables z and y, since it represents both the relation of a polynomial of the third degree and a polynomial of the fifth degree and the function is not yew. In the case when the variables, in turn, are functions of the variable x: then the function] is called a rational function of the Example functions. A function is a rational function of r and pvdikvlv Line 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 in terms of elementary functions ... For example, integrals often found 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 by some substitutions to the integration of rational functions. 1. Suppose it is required 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 satisfying the condition ad - bc\u003e 0 (for ad - be \u003d 0 the coefficients a and b are proportional to the coefficients c and d, and therefore the ratio does not depend on x; hence, in this case the integrand will be a rational function of the variable x, the integration of which was considered earlier). Let us make a change of variable in this integral by setting Hence, we express the variable x in terms of the new variable. We have x \u003d - a rational function of t. Further, we find or, after simplification, Therefore, where A1 (t) is a rational function of *, since the rational function 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 will be equal to At. IVIt integral 4 The integrand * function is a rational function of. Therefore, we set t \u003d Then Integration of rational functions Brief information about rational functions Integration of elementary fractions General case Integration of irrational functions First Euler substitution Second Euler substitution Third Euler substitution Thus, we obtain Primar 5. Find the integral The common denominator of fractional exponents x is 12, therefore the integrand the function can be represented in the form 1 _ 1_ from which it can be seen that it is a rational function of: Taking this into account, we put. Therefore, 2. Consider intefps of the form where the subintephalic function is such that 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 an integral of a rational function of another variable by Euler's substitutions. 8.1. Euler's first substitution Let the coefficient a\u003e 0. We put or From here we find x as a rational function of and, therefore, Thus, the indicated substitution expresses rationally through *. Therefore, we will have where Remark. The first Euler substitution can also be taken in the form Example 6. Find the integral We will therefore have dx Euler's substitution, show that Y 8. 2. Euler's second substitution Let the trinomial ax2 + bx + c have different real roots λ] and x2 (the coefficient may 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, the original integral is reduced to an integral of a rational function, that is, where Problem. Using the first Euler substitution, show that is a rational function of t. Example 7. Neyti integral dx M function] - x1 has different real roots. Therefore, we apply the second substitution to Euler From here we find Substituting the found cutouts into Given? In * gyvl; we get 8.3. Euler's third substation Let the coefficient c\u003e 0. Change the variable by setting. Note that the first and second Euler substitutions are sufficient to reduce the integral to the integral of a rational function. Indeed, if the discriminant b2 -4ac\u003e 0, then the roots of the square 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\u003e 0. In this case, the first Euler substitution is applicable. To find integrals of the type indicated above, it is not always advisable to use Euler 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, select the long square from the square of the th trinomial: where After that, make a substitution and get where the coefficients a and P have different signs or they are both positive. For, and also for a\u003e 0 and the integral will be reduced to the logarithm, if, on the other hand, to the arcsine. When. Find imtegrel 4 So something like that. assuming we obtain Prmmar 9. Find. We set x -, we will have 2. An integral of the form reduces to the integral y from item 1 as follows. Taking into account that the derivative () "\u003d 2, we select it in the numerator: 4 We identify the derivative of the radical expression in the numerator. Since (x, we will have, taking into account the result of Example 9, 3. Integrals of the form where P„ (x) is the polynomial n -th degree, can be found by the method of indefinite coefficients, which consists in the following: Suppose that the equality holds Example 10. Mighty integral where Qn-i (s) is a polynomial of (n - 1) -th degree with indefinite coefficients: To find the unknowns coefficients | we differentiate both sides of (1): Then the right-hand side of equality (2) is reduced to a common denominator equal to the denominator of the left-hand side, i.e. y / ax2 + bx + c, canceling both sides of (2) by which, we obtain an identity in on both sides of which there are polynomials of degree n. Equating the coefficients at the same powers of x in the left and right sides of (3), we obtain n + 1 equations, from which we find the required coefficients j4 * (fc \u003d 0,1,2, ..., n). Substituting their values \u200b\u200binto the right-hand side of (1) and finding the integral + с, we obtain the answer for this integral. Example 11. Find the integral We put Differentiating both suits of equality, we will have. Bringing the right side to a common denominator and canceling 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 \u003d Then we find the integral on the right-hand side of equality (4): Therefore, the required integral will be equal to

Integrals of the form (m 1, n 1, m 2, n 2,… are integers). In these integrals, the integrand is rational with respect to the variable of integration and radicals of x. They are calculated by substituting x \u003d t s, where s is the common denominator of fractions, ... With this change of variable, all ratios \u003d r 1, \u003d r 2, ... are integers, that is, the integral is reduced to a rational function of the variable t:

Integrals of the form (m 1, n 1, m 2, n 2,… are integers). These integrals are substituted:

where s is the common denominator of fractions,…, are reduced to a rational function of the variable t.

Integrals of the form To calculate the integral I 1, a complete square is allocated under the radical sign:

and the substitution is applied:

As a result, this integral is reduced to a tabular one:

In the numerator of the integral I 2, the differential of the expression under the radical sign is distinguished, and this integral is represented as the sum of two integrals:

where I 1 is the integral calculated above.

The calculation of the integral I 3 is reduced to the calculation of the integral I 1 by the substitution:

Integral of the form Particular cases of calculating integrals of this form were considered in the previous section. There are several different ways to calculate them. Consider one of these techniques, based on the use of trigonometric substitutions.

The square trinomial ax 2 + bx + c by selecting a complete square and changing the variable can be represented in the form Thus, it is enough to limit ourselves to considering three types of integrals:

Integral substitution

u \u003d ksint (or u \u003d kcost)

reduces to an integral of a rational function with respect to sint and cost.

Integrals of the form (m, n, p є Q, a, b є R). The considered integrals, called integrals of the differential binomial, are expressed in terms of elementary functions only in the following three cases:

1) if p є Z, then the substitution is applied:

where s is the common denominator of the fractions m and n;

2) if Z, then the substitution is used:

where s is the denominator of the fraction

3) if Z, then the substitution is applied:

where s is the denominator of the fraction

The online calculator serves to calculate integrals of irrational fractions of the form,,.

Let be - rational function of This function, and hence the integral of it, is rationalized by the substitution x \u003d t r, where r is the least common multiple of the numbers r 1, r 2,…, r n. Then dx \u003d rt r -1 and the integral is a rational function of t. Similarly, if the integrand there is a rational function of , then the integrand is rationalized by substitution where t is the least common multiple of r 1, r 2,…, r n. Then Substituting into the original expression, we obtain a rational function of t.

An example. Calculate. The least common multiple of 2 and 3 is 6. Therefore, we make the change x \u003d t 6. Then dx \u003d 6t 5 dt and

Integration of irrational functions

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

There are times when it is appropriate to use the method of summing up the differential sign. For example, when finding indefinite integrals of the form, where p Is a rational fraction.

Example.

Find the indefinite integral .

Decision.

It's not hard to see that. Therefore, we bring 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, the substitution implies that and

that is, x and dx are expressed in terms of rational functions of t. Moreover, each power of the fraction is expressed in terms of a rational function of t.

Example 33.4... Find the integral

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

Therefore, we put x + 2 \u003d t 6, x \u003d t 6 -2, dx \u003d 6t 5 dt, Therefore,

Example 33.5. Specify a substitution for finding integrals:

Solution: For I 1 the substitution x \u003d t 2, for I 2 the substitution

14. Trigonometric substitution

Integrals of the type are reduced to integrals of functions that rationally depend on trigonometric functions using the following trigonometric substitutions: x \u003d a sint for the first integral; x \u003d a tgt for the second integral; for the third integral.

Example 33.6. Find the integral

Solution: Put x \u003d 2 sin t, dx \u003d 2 cos tdt, t \u003d arcsin x / 2. Then

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

Example 33.7. Find the integral

Solution: Since x 2 + 2x-4 \u003d (x + 1) 2 -5, then x + 1 \u003d t, x \u003d t-1, dx \u003d dt. therefore We put

Note: Integral of type it is expedient to find using the substitution x \u003d 1 / t.

15. Definite integral

Let the function be given on a segment and has an antiderivative on it. The difference is called definite integral functions by segment and denote. So,

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

For example, one of the antiderivatives for a function. therefore

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

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

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

Then It follows from this that the function c (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 it is integrable on [a; b] their sum u

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

▼
▲

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

3.

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

4. If the function (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 since the limit of the integral sum is independent of the method of dividing the segment [a; b] into parts). If c \u003d 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 arrangement of points a, b, c (we assume that the function (x) is integrable on the largest of the resulting intervals).

So, for example, if a< b < с, то

(properties 4 and 3 are used).

5. "The mean theorem". If the function (x) is continuous on the segment [a; b], then there is a thin c є [a; b] such that

▼ By the Newton-Leibniz formula, we have

where F "(x) \u003d ƒ (x). Applying to the difference F (b) -F (a) the Lagrange theorem (the theorem on the finite increment of a function), we obtain

F (b) -F (a) \u003d F "(c) (b-a) \u003d ƒ (c) (b-a). ▲

Property 5 ("mean value theorem") for ƒ (x) ≥ 0 has a simple geometric meaning: the value of the definite integral is, for some c є (a; b), the area of \u200b\u200ba rectangle with height ƒ (c) and base b-a (see Fig. 170). Number

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

6. If the function ƒ (x) preserves 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 Î [a; b], then

ƒ (c) ≥0, b-a\u003e 0.

Therefore, ƒ (c) (b-a) ≥ 0, i.e. ▲

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

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

Or, according to property 2,

Note that inequalities cannot be differentiated.

8. Estimation of the integral. If m and M are respectively the smallest and largest values \u200b\u200bof the function y \u003d 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 \u200b\u200ba curvilinear trapezoid is enclosed between the areas of rectangles, the base of which is, and the heights are equal to 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

Hence 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 variable of integration is replaced by this limit, that is,

Calculating the area of \u200b\u200ba figure is one of the most difficult problems in the theory of areas. 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 the calculation of the areas of more complex shapes. When solving such problems, one has to resort to integral calculus.

In this article, we will consider the problem of calculating the area of \u200b\u200ba curved trapezoid, and we will approach it in a geometric sense. This will allow us to find out the direct relationship between a definite integral and the area of \u200b\u200ba curved trapezoid.

Let the function y \u003d f (x) continuous on the segment and does not change sign on it (that is, non-negative or non-positive). Figure Gbounded by lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d bare called curved trapezoid... Let us denote its area S (G).

Let us approach the problem of calculating the area of \u200b\u200ba curved trapezoid as follows. In the squared figures section, we found out that a curved trapezoid is a squared figure. If you split the segment on n parts by dots and designate , and the points should be chosen so that for, then the figures corresponding to the lower and upper Darboux sums can be considered as entering P and encompassing Q polygonal shapes for G.

Thus, with an increase in the number of split points n, we arrive at the inequality, where is an arbitrarily small positive number, and s and S - the lower and upper Darboux sums for a given segment partition ... In another entry ... Therefore, turning to the concept of a definite Darboux integral, we obtain .

The last equality means that a definite integral for a continuous and non-negative function y \u003d f (x) represents in the geometric sense the area of \u200b\u200bthe corresponding curved trapezoid. This is geometric meaning of definite integral.

That is, calculating a definite integral, we find the area of \u200b\u200bthe figure bounded by the lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d b.

Comment.

If the function y \u003d f (x) non-positive on the segment , then the area of \u200b\u200ba curved trapezoid can be found as .

Example.

Calculate the area of \u200b\u200ba shape bounded by lines .

Decision.

Let's build a figure on a plane: straight y \u003d 0 coincides with the abscissa, straight x \u003d -2and x \u003d 3 are parallel to the ordinate axis, and the curve can be plotted using geometric transformations of the function graph.

Thus, we need to find the area of \u200b\u200ba curved trapezoid. The geometric meaning of a definite integral indicates to us that the required area is expressed by a definite integral. Hence, ... This definite integral can be calculated using the Newton-Leibniz formula.