Integration of the simplest irrational fractions. Integration of irrational expressions. Integrating Complex Fractions

Definition 1

The set of all antiderivatives of a given function $y=f(x)$, defined on a certain segment, is called the indefinite integral of a given function $y=f(x)$. The indefinite integral is denoted by the symbol $\int f(x)dx $.

Comment

Definition 2 can be written as follows:

\[\int f(x)dx =F(x)+C.\]

Not every irrational function can be expressed as an integral through elementary functions. However, most of these integrals can be reduced using substitutions to integrals of rational functions, which can be expressed in terms of elementary functions.

$\int R\left(x,x^(m/n) ,...,x^(r/s) \right)dx $;

$\int R\left(x,\left(\frac(ax+b)(cx+d) \right)^(m/n) ,...,\left(\frac(ax+b)(cx +d) \right)^(r/s) \right)dx $;

$\int R\left(x,\sqrt(ax^(2) +bx+c) \right)dx $.

I

When finding an integral of the form $\int R\left(x,x^(m/n) ,...,x^(r/s) \right)dx $ it is necessary to perform the following substitution:

With this substitution, each fractional power of the variable $x$ is expressed through an integer power of the variable $t$. As a result, the integrand function is transformed into a rational function of the variable $t$.

Example 1

Perform integration:

\[\int \frac(x^(1/2) dx)(x^(3/4) +1) .\]

Solution:

$k=4$ is the common denominator of the fractions $\frac(1)(2) ,\, \, \frac(3)(4) $.

\ \[\begin(array)(l) (\int \frac(x^(1/2) dx)(x^(3/4) +1) =4\int \frac(t^(2) ) (t^(3) +1) \cdot t^(3) dt =4\int \frac(t^(5) )(t^(3) +1) dt =4\int \left(t^( 2) -\frac(t^(2) )(t^(3) +1) \right)dt =4\int t^(2) dt -4\int \frac(t^(2) )(t ^(3) +1) dt =\frac(4)(3) \cdot t^(3) -) \\ (-\frac(4)(3) \cdot \ln |t^(3) +1 |+C)\end(array)\]

\[\int \frac(x^(1/2) dx)(x^(3/4) +1) =\frac(4)(3) \cdot \left+C\]

II

When finding an integral of the form $\int R\left(x,\left(\frac(ax+b)(cx+d) \right)^(m/n) ,...,\left(\frac(ax+ b)(cx+d) \right)^(r/s) \right)dx $ it is necessary to perform the following substitution:

where $k$ is the common denominator of the fractions $\frac(m)(n) ,...,\frac(r)(s) $.

As a result of this substitution, the integrand function is transformed into a rational function of the variable $t$.

Example 2

Perform integration:

\[\int \frac(\sqrt(x+4) )(x) dx .\]

Solution:

Let's make the following substitution:

\ \[\int \frac(\sqrt(x+4) )(x) dx =\int \frac(t^(2) )(t^(2) -4) dt =2\int \left(1 +\frac(4)(t^(2) -4) \right)dt =2\int dt +8\int \frac(dt)(t^(2) -4) =2t+2\ln \left |\frac(t-2)(t+2) \right|+C\]

After making the reverse substitution, we get the final result:

\[\int \frac(\sqrt(x+4) )(x) dx =2\sqrt(x+4) +2\ln \left|\frac(\sqrt(x+4) -2)(\ sqrt(x+4) +2) \right|+C.\]

III

When finding an integral of the form $\int R\left(x,\sqrt(ax^(2) +bx+c) \right)dx $, the so-called Euler substitution is performed (one of three possible substitutions is used).

Euler's first substitution

For the case $a>

Taking the “+” sign in front of $\sqrt(a) $, we get

Example 3

Perform integration:

\[\int \frac(dx)(\sqrt(x^(2) +c) ) .\]

Solution:

Let's make the following substitution (case $a=1>0$):

\[\sqrt(x^(2) +c) =-x+t,\, \, x=\frac(t^(2) -c)(2t) ,\, \, dx=\frac(t ^(2) +c)(2t^(2) ) dt,\, \, \sqrt(x^(2) +c) =-\frac(t^(2) -c)(2t) +t= \frac(t^(2) +c)(2t) .\] \[\int \frac(dx)(\sqrt(x^(2) +c) ) =\int \frac(\frac(t^ (2) +c)(2t^(2) ) dt)(\frac(t^(2) +c)(2t) ) =\int \frac(dt)(t) =\ln |t|+C \]

After making the reverse substitution, we get the final result:

\[\int \frac(dx)(\sqrt(x^(2) +c) ) =\ln |\sqrt(x^(2) +c) +x|+C.\]

Euler's second substitution

For the case $c>0$ it is necessary to perform the following substitution:

Taking the “+” sign in front of $\sqrt(c) $, we get

Example 4

Perform integration:

\[\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x+x^(2) ) ) dx .\]

Solution:

Let's make the following substitution:

\[\sqrt(1+x+x^(2) ) =xt+1.\]

\ \[\sqrt(1+x+x^(2) ) =xt+1=\frac(t^(2) -t+1)(1-t^(2) ) \] \

$\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x+x^(2) ) ) dx = \int \frac((-2t^(2) +t)^(2) (1-t)^(2) (1-t^(2))(2t^(2) -2t+2))( (1-t^(2))^(2) (2t-1)^(2) (t^(2) -t+1)(1-t^(2))^(2) ) dt =\ int \frac(t^(2) )(1-t^(2) ) dt =-2t+\ln \left|\frac(1+t)(1-t) \right|+C$ Having made the reverse substitution, we get the final result:

\[\begin(array)(l) (\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x +x^(2) ) dx =-2\cdot \frac(\sqrt(1+x+x^(2) ) -1)(x) +\ln \left|\frac(x+\sqrt(1 +x+x^(2) ) -1)(x-\sqrt(1+x+x^(2) ) +1) \right|+C=-2\cdot \frac(\sqrt(1+x +x^(2) ) -1)(x) +) \\ (+\ln \left|2x+2\sqrt(1+x+x^(2) ) +1\right|+C) \end (array)\]

Euler's third substitution

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

The concept of a rational function of one variable can be extended to multiple arguments. If for 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, \ dotsc, w)%%. The arguments of such a function can themselves be functions of the independent variable %%x%%, including radicals of the form %%\sqrt[n](x), n \in \mathbb(N)%%. For example, the rational function $$ R(u,v,w) = \frac(u + v^2)(w) $$ with %%u = x, v = \sqrt(x)%% and %%w = \sqrt(x^2 + 1)%% is a rational function of $$ R\left(x, \sqrt(x), \sqrt(x^2+1)\right) = \frac(x + \sqrt(x ^2))(\sqrt(x^2 + 1)) = f(x) $$ 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%%.

Let's consider integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%%. Such integrals are rationalized by replacing the variable %%t = \sqrt[n](x)%%, then %%x = t^n, \mathrm(d)x = nt^(n-1)%%.

Example 1

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

The integrand of the desired argument is written as a function of radicals of degree %%2%% and %%3%%. Since the least common multiple of %%2%% and %%3%% is %%6%%, this integral is an integral of type %%\int R(x, \sqrt(x)) \mathrm(d)x %% and can be rationalized by replacing %%\sqrt(x) = t%%. Then %%x = t^6, \mathrm(d)x = 6t \mathrm(d)t, \sqrt(x) = t^3, \sqrt(x) =t^2%%. Therefore, $$ \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) = \int \frac(6t^5 \mathrm(d)t)(t^3 + t^2) = 6\int\frac(t^3)(t+1)\mathrm(d)t. $$ Let's take %%t + 1 = z, \mathrm(d)t = \mathrm(d)z, z = t + 1 = \sqrt(x) + 1%% and $$ \begin(array)(ll ) \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) &= 6\int\frac((z-1)^3)(z) \mathrm(d) t = \\ &= 6\int z^2 dz -18 \int z \mathrm(d)z + 18\int \mathrm(d)z -6\int\frac(\mathrm(d)z)(z ) = \\ &= 2z^3 - 9 z^2 + 18z -6\ln|z| + C = \\ &= 2 \left(\sqrt(x) + 1\right)^3 - 9 \left(\sqrt(x) + 1\right)^2 + \\ &+~ 18 \left( \sqrt(x) + 1\right) - 6 \ln\left|\sqrt(x) + 1\right| + C \end(array) $$

Integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%% are a special case of fractional linear 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 replacing the variable %%t = \sqrt[n](\dfrac(ax+b)(cd+d))%%, then %%x = \dfrac(dt^n - b)(a - ct^n)%%. Then $$ \mathrm(d)x = \frac(n t^(n-1)(ad - bc))(\left(a - ct^n\right)^2)\mathrm(d)t. $$

Example 2

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

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

Let's 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 ones if, after isolating the complete square, a change of variables is made.

Example 3

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

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

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

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 highlight 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.

3.

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 were 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.

Let the function y = f(x) continuous on the segment and does not change the sign on it (that is, non-negative or non-positive). Figure G, bounded by lines y = f(x), y = 0, x = a And x = b, called curved trapezoid. Let's denote its area S(G).

Let us approach the problem of calculating the area of ​​a curvilinear trapezoid as follows. In the section on squarable figures, we found out that a curved trapezoid is a squarable figure. If you split the segment on n parts with dots to indicate , and choose points so that for , then the figures corresponding to the lower and upper Darboux sums can be considered included P and comprehensive Q polygonal shapes for G.

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

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

That is, by calculating the definite integral, we will find the area of ​​the figure bounded by the lines y = f(x), y = 0, x = a And x = b.

Comment.

If the function y = f(x) non-positive on the segment , then the area of ​​a curved trapezoid can be found as .

Example.

Calculate the area of ​​a figure bounded by lines .

Solution.

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

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

Integrals of the form (m 1, n 1, m 2, n 2, ... - integers). In these integrals, the integrand is rational with respect to the integration variable and the radicals of x. They are calculated by substituting x=t s, where s is the common denominator of the fractions, ... With such a replacement of the variable, all relations = r 1, = r 2, ... are integers, i.e. the integral is reduced to a rational function of the variable t:

Integrals of the form (m 1, n 1, m 2, n 2, ... - integers). These integrals are by substitution:

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

Integrals of the form To calculate the integral I 1, select a complete square 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 highlighted, 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 substitution:

Integral of the form Special cases of calculating integrals of this type are considered in the previous paragraph. There are several different methods for calculating them. Let's consider one of these techniques, based on the use of trigonometric substitutions.

The square trinomial ax 2 +bx+c by isolating the 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 by substitution

u=ksint (or u=kcost)

reduces to the 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 integrals under consideration, called integrals of a differential binomial, are expressed through 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 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 basic methods for integrating irrational functions (roots) are given. They include: integration of linear fractional irrationality, differential binomial, integrals with the square root of a square trinomial. Trigonometric substitutions and Euler substitutions are given. Some elliptic integrals expressed through elementary functions are considered.

Content

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, 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: 0 Let's make a substitution. 0 Here at x >< 0 (u>< 0 ) take the upper sign ′+ ′. At x

.

(u
) - lower ′- ′.

References: