Integration of simple fractions of 3 types. Integration of rational fractions. Decomposition of a proper rational fraction into a sum of simple fractions

Enter the function for which you want to find the integral

After calculating the indefinite integral, you can get free DETAILED solution the integral you entered.

Let's find the solution of the indefinite integral of the function f(x) (the antiderivative of the function).

Examples

With the use of degree
(square and cube) and fractions

(x^2 - 1)/(x^3 + 1)

Square root

Sqrt(x)/(x + 1)

cube root

Cbrt(x)/(3*x + 2)

Using sine and cosine

2*sin(x)*cos(x)

Arcsine

X*arcsin(x)

Arc cosine

x*arccos(x)

Application of the logarithm

X*log(x, 10)

natural logarithm

Exhibitor

Tg(x)*sin(x)

Cotangent

Ctg(x)*cos(x)

Irrational fractions

(sqrt(x) - 1)/sqrt(x^2 - x - 1)

Arctangent

X*arctg(x)

Arc tangent

X*arсctg(x)

Hyberbolic sine and cosine

2*sh(x)*ch(x)

Hyberbolic tangent and cotangent

ctgh(x)/tgh(x)

Hyberbolic arcsine and arccosine

X^2*arcsinh(x)*arccosh(x)

Hyberbolic arctangent and arccotangent

X^2*arctgh(x)*arctgh(x)

Rules for entering expressions and functions

Expressions can consist of functions (notations are given in alphabetical order): absolute(x) Absolute value x
(module x or |x|) arccos(x) Function - arc cosine of x arccosh(x) Arc cosine hyperbolic from x arcsin(x) Arcsine from x arcsinh(x) Arcsine hyperbolic from x arctg(x) Function - arc tangent from x arctgh(x) The arc tangent is hyperbolic from x e e a number that is approximately equal to 2.7 exp(x) Function - exponent from x(which is e^x) log(x) or log(x) Natural logarithm of x
(To obtain log7(x), you need to enter log(x)/log(7) (or, for example, for log10(x)=log(x)/log(10)) pi The number is "Pi", which is approximately equal to 3.14 sin(x) Function - Sine of x cos(x) Function - Cosine of x sinh(x) Function - Hyperbolic sine of x cash(x) Function - Hyperbolic cosine of x sqrt(x) Function - Square root from x sqr(x) or x^2 Function - Square x tg(x) Function - Tangent from x tgh(x) Function - Hyperbolic tangent of x cbrt(x) The function is the cube root of x

You can use the following operations in expressions: Real numbers enter in the form 7.5 , not 7,5 2*x- multiplication 3/x- division x^3- exponentiation x + 7- addition x - 6- subtraction
Other features: floor(x) Function - rounding x down (example floor(4.5)==4.0) ceiling(x) Function - rounding x up (example ceiling(4.5)==5.0) sign(x) Function - Sign x erf(x) Error function (or probability integral) laplace(x) Laplace function

The problem of finding the indefinite integral of a fractionally rational function is reduced to integrating simple fractions. Therefore, we recommend that you first familiarize yourself with the section on the theory of decomposition of fractions into simple ones.

Example.

Find the indefinite integral.

Solution.

Since the degree of the numerator of the integrand is equal to the degree of the denominator, first we select the integer part by dividing the polynomial by the polynomial by the column:

That's why, .

The decomposition of the obtained proper rational fraction into simple fractions has the form . Consequently,

The resulting integral is an integral of the simplest fraction of the third type. Looking ahead a little, we note that it can be taken by bringing it under the differential sign.

Because , then . That's why

Consequently,

Now let's move on to describing the methods for integrating the simplest fractions of each of the four types.

Integration of the simplest fractions of the first type

The method of direct integration is ideal for solving this problem:

Example.

Find the set of antiderivatives of a function

Solution.

Let's find the indefinite integral using the properties of the antiderivative, the table of antiderivatives and the integration rule.

Top of page

Integration of the simplest fractions of the second type

The method of direct integration is also suitable for solving this problem:

Example.

Solution.

Top of page

Integration of the simplest fractions of the third type

First, we present the indefinite integral as a sum:

We take the first integral by the method of subsuming under the sign of the differential:

That's why,

We transform the denominator of the resulting integral:

Consequently,

The formula for integrating the simplest fractions of the third type takes the form:

Example.

Find the indefinite integral .

Solution.

We use the resulting formula:

If we didn't have this formula, what would we do:

Top of page

Integration of the simplest fractions of the fourth type

The first step is to sum it up under the differential sign:

The second step is to find an integral of the form . Integrals of this type are found using recurrent formulas. (See section integrating using recursive formulas). For our case, the following recursive formula is suitable:

Example.

Find the indefinite integral

Solution.

For this type of integrand, we use the substitution method. Let us introduce a new variable (see the section on integration and rational functions):

After substitution we have:

We came to finding the integral of a fraction of the fourth type. In our case, we have the coefficients M=0, p=0, q=1, N=1 And n=3. We apply the recursive formula:

After the reverse substitution, we get the result:

Integration trigonometric functions

1. Integrals of the form

are calculated by converting the product of trigonometric functions into a sum according to the formulas:

For example, 2. Integrals of the form

, where m or n- an odd positive number, are calculated by subsuming under the sign of the differential. For example,

3. Integrals of the form

, where m And n- even positive numbers, are calculated using the reduction formulas: For example,

4. Integrals

where are calculated by changing the variable: or For example,

5. Integrals of the form are reduced to integrals of rational fractions using the universal trigonometric substitution then (because

=[after dividing the numerator and denominator by ]= ;

For example,

It should be noted that the use of a universal substitution often leads to cumbersome calculations.

§five. Integration of the simplest irrationalities

Consider methods for integrating the simplest types of irrationality. one.

Functions of this type are integrated in the same way as the simplest rational fractions of the 3rd type: in the denominator from the square trinomial, full square and a new variable is introduced. Example.

(under the sign of the integral is the rational function of the arguments). Integrals of this kind are calculated using the substitution . In particular, in integrals of the form we denote . If the integrand contains roots different degrees:

, then denote , where n is the least common multiple of the numbers m,k. Example 1

Example 2

is an improper rational fraction, select the integer part:

–

3. Integrals of the form are calculated using trigonometric substitutions:

45 Definite Integral

Definite integral is an additive monotone normalized functional defined on a set of pairs, the first component of which is an integrable function or functional, and the second is an area in the set of this function (functional).

Definition

Let it be defined on . Let's break it into parts with several arbitrary points. Then we say that the segment has been partitioned Next, we choose an arbitrary point , ,

The definite integral of a function on a segment is the limit of integral sums as the partition rank tends to zero, if it exists regardless of the partition and choice of points, that is

If this limit exists, then the function is said to be Riemann integrable on.

Notation

· - lower limit.

· - upper limit.

· - integrand function.

· - length of a partial segment.

· is the integral sum of the function on the corresponding partition .

· - maximum length of a partial segment.

Properties

If a function is Riemann-integrable on , then it is bounded on it.

geometric sense

The definite integral as the area of a figure

Definite integral numerically equal to area figure bounded by the x-axis, straight lines and and function graph.

Newton-Leibniz theorem

[edit]

(redirected from "Newton-Leibniz Formula")

Newton - Leibniz formula or fundamental theorem of analysis gives the relation between two operations: taking a definite integral and calculating an antiderivative.

Proof

Let an integrable function be given on the segment. Let's start by noting that

that is, it does not matter which letter ( or ) is under the sign in a definite integral over the interval .

Set an arbitrary value and define a new function . It is defined for all values of , because we know that if there is an integral of on , then there is also an integral of on , where . Recall that we consider by definition

(1)

notice, that

Let us show that it is continuous on the segment . Indeed, let ; then

and if , then

Thus, is continuous on regardless of whether it has discontinuities or not; it is important that it is integrable on .

The figure shows a graph. The area of the variable figure is . Its increment is equal to the area of the figure , which, due to the boundedness of , obviously tends to zero at regardless of whether it is a point of continuity or discontinuity, for example, a point .

Now let the function not only be integrable on , but be continuous at the point . Let us prove that then has a derivative at this point equal to

(2)

Indeed, for the given point

(1) , (3)

We put , and since the constant is relative to ,TO . Further, due to the continuity at the point, for any one can specify such that for .

which proves that the left side of this inequality is o(1) for .

Passing to the limit in (3) at shows the existence of the derivative of at the point and the validity of equality (2). Here we are talking about the right and left derivatives, respectively.

If a function is continuous on , then, based on what was proved above, the corresponding function

(4)

has a derivative equal to . Therefore, the function is antiderivative for on .

This conclusion is sometimes called the Variable Upper Limit Integral Theorem or Barrow's Theorem.

We have proved that an arbitrary continuous function on an interval has an antiderivative on this interval, defined by equality (4). This proves the existence of an antiderivative for any function continuous on an interval.

Now let there be an arbitrary antiderivative of a function on the . We know that , where is some constant. Assuming in this equality and taking into account that , we obtain .

In this way, . But

Improper integral

[edit]

From Wikipedia, the free encyclopedia

Definite integral called improper if at least one of the following conditions is true:

· The limit a or b (or both limits) is infinite;

· The function f(x) has one or more breakpoints inside the segment .

[edit] Improper integrals of the first kind

. Then:

1. If and the integral is called . In this case is called convergent.

, or simply divergent.

Let be defined and continuous on the set from and . Then:

1. If , then the notation and the integral is called improper Riemann integral of the first kind. In this case is called convergent.

2. If there is no final ( or ), then the integral is said to be divergent to , or simply divergent.

If the function is defined and continuous on the entire real line, then there may be an improper integral of this function with two infinite limits of integration, which is determined by the formula:

, where c is an arbitrary number.

[edit] The geometric meaning of the improper integral of the first kind

The improper integral expresses the area of an infinitely long curvilinear trapezoid.

[edit] Examples

[edit] Improper integrals of the second kind

Let it be defined on , suffer an infinite discontinuity at the point x=a and . Then:

1. If , then the notation and the integral is called

is called divergent to , or simply divergent.

Let it be defined on , suffer an infinite discontinuity at x=b and . Then:

1. If , then the notation and the integral is called improper Riemann integral of the second kind. In this case, the integral is called convergent.

2. If or , then the designation is preserved, and is called divergent to , or simply divergent.

If the function suffers a discontinuity at an internal point of the segment, then the improper integral of the second kind is determined by the formula:

[edit] Geometric meaning of improper integrals of the second kind

The improper integral expresses the area of an infinitely high curvilinear trapezoid

[edit] Example

[edit] Special case

Let the function be defined on the entire real axis and have a discontinuity at points .

Then we can find the improper integral

[edit] Cauchy criterion

1. Let be defined on the set from and .

Then converges

2. Let is defined on and .

Then converges

[edit] Absolute convergence

Integral called absolutely convergent, if converges.
If an integral converges absolutely, then it converges.

[edit] Conditional convergence

The integral is called conditionally convergent if converges and diverges.

48 12. Improper integrals.

When considering definite integrals, we assumed that the region of integration is bounded (more specifically, it is the segment [ a ,b ]); for the existence of a definite integral, the boundedness of the integrand on [ a ,b ]. We will call definite integrals for which both these conditions are satisfied (boundedness of both the integration domain and the integrand) own; integrals for which these requirements are violated (i.e., either the integrand, or the domain of integration, or both, is unbounded) non-own. In this section, we will study improper integrals.

12.1. Improper integrals over an unbounded interval (improper integrals of the first kind).

12.1.1. Definition of an improper integral over an infinite interval. Examples.
12.1.2. The Newton-Leibniz formula for the improper integral.
12.1.3. Comparison criteria for non-negative functions.

12.1.3.1. Sign of comparison.
12.1.3.2. A sign of comparison in the limiting form.

12.1.4. Absolute convergence of improper integrals over an infinite interval.
12.1.5. Convergence criteria for Abel and Dirichlet.

12.2. Improper integrals of unbounded functions (improper integrals of the second kind).

12.2.1. Definition of an improper integral of an unbounded function.

12.2.1.1. Singularity at the left end of the interval of integration.
12.2.1.2. Application of the Newton-Leibniz formula.
12.2.1.3. Singularity at the right end of the interval of integration.
12.2.1.4. Singularity at an interior point of the integration interval.
12.2.1.5. Several singularities on the interval of integration.

12.2.2. Comparison criteria for non-negative functions.

12.2.2.1. Sign of comparison.
12.2.2.2. A sign of comparison in the limiting form.

12.2.3. Absolute and conditional convergence of improper integrals of discontinuous functions.
12.2.4. Convergence criteria for Abel and Dirichlet.

12.1. Improper integrals over an unbounded interval

(improper integrals of the first kind).

12.1.1. Definition of an improper integral over an infinite interval. Let the function f (x ) is defined on the half-line and is integrable over any interval [ from, implying in each of these cases the existence and finiteness of the corresponding limits. Now the solutions of the examples look more simple: .

12.1.3. Comparison criteria for non-negative functions. In this section, we will assume that all integrands are non-negative over the entire domain of definition. Until now, we have determined the convergence of the integral by calculating it: if there is a finite limit of the antiderivative with the corresponding aspiration ( or ), then the integral converges, otherwise it diverges. When solving practical problems, however, it is important first of all to establish the very fact of convergence, and only then calculate the integral (besides, the antiderivative is often not expressed in terms of elementary functions). We formulate and prove a number of theorems that allow us to establish the convergence and divergence of improper integrals of nonnegative functions without calculating them.
12.1.3.1. Comparison sign. Let the functions f (x ) And g (x ) integr

TOPIC: Integration of rational fractions.

Attention! When studying one of the main methods of integration - the integration of rational fractions - it is required to consider polynomials in the complex domain for rigorous proofs. Therefore, it is necessary study in advance some properties complex numbers and operations on them.

Integration of the simplest rational fractions.

If P(z) And Q(z) are polynomials in the complex domain, then is a rational fraction. It is called correct if the degree P(z) less degree Q(z) , And wrong if the degree R no less degree Q.

Any improper fraction can be represented as: ,

P(z) = Q(z) S(z) + R(z),

a R(z) – polynomial whose degree is less than the degree Q(z).

Thus, the integration of rational fractions is reduced to the integration of polynomials, that is, power functions, and proper fractions, since it is a proper fraction.

Definition 5. The simplest (or elementary) fractions are fractions of the following types:

1) , 2) , 3) , 4) .

Let's find out how they are integrated.

3) (explored earlier).

Theorem 5. Any proper fraction can be represented as a sum of simple fractions (without proof).

Corollary 1. If is a proper rational fraction, and if among the roots of the polynomial there are only simple real roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 1st type:

Example 1

Corollary 2. If is a proper rational fraction, and if there are only multiple real roots among the roots of the polynomial, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 1st and 2nd types:

Example 2

Corollary 3. If is a proper rational fraction, and if among the roots of the polynomial there are only simple complex conjugate roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 3rd type:

Example 3

Corollary 4. If is a proper rational fraction, and if among the roots of the polynomial there are only multiple complex conjugate roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 3rd and 4th types:

To determine the unknown coefficients in the above expansions, proceed as follows. The left and right parts of the expansion containing unknown coefficients are multiplied by The equality of two polynomials is obtained. Equations for the desired coefficients are obtained from it, using that:

1. equality is valid for any values of X (method of partial values). In this case, any number of equations are obtained, any m of which allow us to find unknown coefficients.

2. the coefficients coincide at the same powers of X (method of indefinite coefficients). In this case, a system of m - equations with m - unknowns is obtained, from which unknown coefficients are found.

3. combined method.

Example 5. Expand a fraction to the simplest.

Solution:

Find the coefficients A and B.

1 way - private value method:

Method 2 - the method of uncertain coefficients:

Answer:

Integration of rational fractions.

Theorem 6. The indefinite integral of any rational fraction on any interval on which its denominator is not equal to zero exists and is expressed in terms of elementary functions, namely rational fractions, logarithms and arctangents.

Proof.

We represent a rational fraction in the form: . Moreover, the last term is a proper fraction, and by Theorem 5 it can be represented as a linear combination of simple fractions. Thus, integrating a rational fraction reduces to integrating a polynomial S(x) and the simplest fractions, whose antiderivatives, as was shown, have the form indicated in the theorem.

Comment. The main difficulty in this case is the decomposition of the denominator into factors, that is, the search for all its roots.

Example 1. Find the integral

The integrand is a proper rational fraction. The expansion into irreducible factors of the denominator has the form This means that the expansion of the integrand into the sum of simple fractions has the following form:

Let us find the expansion coefficients by the combined method:

In this way,

Example 2. Find the integral

The integrand is an improper fraction, so we select the integer part:

The first of the integrals is tabular, and the second is calculated by expanding the proper fraction into simple ones:

We have by the method of indefinite coefficients:

In this way,

The derivation of formulas for calculating integrals from the simplest, elementary, fractions of four types is given. More complex integrals, from fractions of the fourth type, are calculated using the reduction formula. An example of integration of a fraction of the fourth type is considered.

Content

See also: Table of indefinite integrals
Methods for calculating indefinite integrals

As is known, any rational function of some variable x can be decomposed into a polynomial and simple, elementary, fractions. There are four types of simple fractions:
1) ;
2) ;
3) ;
4) .
Here a, A, B, b, c are real numbers. Equation x 2+bx+c=0 does not have real roots.

Integration of fractions of the first two types

Integration of the first two fractions is done using the following formulas from the table of integrals:
,
, n ≠ - 1 .

1. Integration of a fraction of the first type

A fraction of the first type by substitution t = x - a is reduced to a table integral:
.

2. Integration of a fraction of the second type

A fraction of the second type is reduced to a table integral by the same substitution t \u003d x - a:

.

3. Integration of a fraction of the third type

Consider the integral of a fraction of the third type:
.
We will calculate it in two steps.

3.1. Step 1. Select the derivative of the denominator in the numerator

We select the derivative of the denominator in the numerator of the fraction. Denote: u = x 2+bx+c. Differentiate: u′ = 2 x + b. Then
;
.
But
.
We omitted the modulo sign because .

Then:
,
where
.

3.2. Step 2. Calculate the integral with A = 0, B=1

Now we calculate the remaining integral:
.

We bring the denominator of the fraction to the sum of squares:
,
where .
We believe that the equation x 2+bx+c=0 has no roots. That's why .

Let's make a substitution
,
.
.

So,
.

Thus, we have found an integral of a fraction of the third type:

,
where .

4. Integration of a fraction of the fourth type

And finally, consider the integral of a fraction of the fourth type:
.
We calculate it in three steps.

4.1) We select the derivative of the denominator in the numerator:
.

4.2) Calculate the integral
.

4.3) Calculate integrals
,
using the cast formula:
.

4.1. Step 1. Extraction of the derivative of the denominator in the numerator

We select the derivative of the denominator in the numerator, as we did in . Denote u = x 2+bx+c. Differentiate: u′ = 2 x + b. Then
.

.
But
.

Finally we have:
.

4.2. Step 2. Calculation of the integral with n = 1

We calculate the integral
.
Its calculation is set out in .

4.3. Step 3. Derivation of the reduction formula

Now consider the integral
.

We present square trinomial to the sum of squares:
.
Here .
We make a substitution.
.
.

We perform transformations and integrate by parts.

.

Multiply by 2(n - 1):
.
We return to x and I n .
,
;
;
.

So, for I n we got the reduction formula:
.
Applying this formula successively, we reduce the integral I n to I 1 .

Example

Calculate Integral

1. We select the derivative of the denominator in the numerator.
;
;

.
Here
.

2. We calculate the integral of the simplest fraction.

.

3. We apply the reduction formula:

for the integral .
In our case b = 1 , c = 1 , 4 c - b 2 = 3. We write out this formula for n = 2 and n = 3 :
;
.
From here

.

Finally we have:

.
We find the coefficient at .
.