The fraction is called correct, if the highest degree of the numerator is less than the highest degree of the denominator. Integral correct rational fraction has the form:
$$ \int \frac(mx+n)(ax^2+bx+c)dx $$
The formula for integrating rational fractions depends on the roots of the polynomial in the denominator. If the polynomial $ ax^2+bx+c $ has:
- Only complex roots, then it is necessary to extract a complete square from it: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(mx+n)(x^2 \pm a ^2) $$
- Different real roots of $ x_1 $ and $ x_2 $, then you need to expand the integral and find uncertain odds$ A $ and $ B $: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)(x-x_1) dx + \int \frac(B )(x-x_2) dx $$
- One multiple root $ x_1 $, then we expand the integral and find the indefinite coefficients $ A $ and $ B $ for the following formula: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)((x-x_1)^2)dx + \int \frac(B)(x-x_1) dx $$
If the fraction is wrong, that is, the highest degree in the numerator is greater than or equal to the highest degree of the denominator, then first it must be reduced to correct form by dividing the polynomial from the numerator by the polynomial from the denominator. IN in this case the formula for integrating a rational fraction has the form:
$$ \int \frac(P(x))(ax^2+bx+c)dx = \int Q(x) dx + \int \frac(mx+n)(ax^2+bx+c)dx $$
Examples of solutions
| Example 1 |
| Find the integral of the rational fraction: $$ \int \frac(dx)(x^2-10x+16) $$ |
| Solution |
|
The fraction is proper and the polynomial has only complex roots. Therefore, we select a complete square: $$ \int \frac(dx)(x^2-10x+16) = \int \frac(dx)(x^2-2\cdot 5 x+ 5^2 - 9) = $$ We fold a complete square and place it under the differential sign $ x-5 $: $$ = \int \frac(dx)((x-5)^2 - 9) = \int \frac(d(x-5))((x-5)^2-9) = $$ Using the table of integrals we obtain: $$ = \frac(1)(2 \cdot 3) \ln \bigg | \frac(x-5 - 3)(x-5 + 3) \bigg | + C = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
| $$ \int \frac(dx)(x^2-10x+16) = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$ |
| Example 2 |
| Perform integration of rational fractions: $$ \int \frac(x+2)(x^2+5x-6) dx $$ |
| Solution |
|
Let's solve the quadratic equation: $$ x^2+5x-6 = 0 $$ $$ x_(12) = \frac(-5\pm \sqrt(25-4\cdot 1 \cdot (-6)))(2) = \frac(-5 \pm 7)(2) $$ We write down the roots: $$ x_1 = \frac(-5-7)(2) = -6; x_2 = \frac(-5+7)(2) = 1 $$ Taking into account the obtained roots, we transform the integral: $$ \int \frac(x+2)(x^2+5x-6) dx = \int \frac(x+2)((x-1)(x+6)) dx = $$ We perform the expansion of a rational fraction: $$ \frac(x+2)((x-1)(x+6)) = \frac(A)(x-1) + \frac(B)(x+6) = \frac(A(x -6)+B(x-1))((x-1)(x+6)) $$ We equate the numerators and find the coefficients $ A $ and $ B $: $$ A(x+6)+B(x-1)=x+2 $$ $$ Ax + 6A + Bx - B = x + 2 $$ $$ \begin(cases) A + B = 1 \\ 6A - B = 2 \end(cases) $$ $$ \begin(cases) A = \frac(3)(7) \\ B = \frac(4)(7) \end(cases) $$ We substitute the found coefficients into the integral and solve it: $$ \int \frac(x+2)((x-1)(x+6))dx = \int \frac(\frac(3)(7))(x-1) dx + \int \frac (\frac(4)(7))(x+6) dx = $$ $$ = \frac(3)(7) \int \frac(dx)(x-1) + \frac(4)(7) \int \frac(dx)(x+6) = \frac(3) (7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$ |
| Answer |
| $$ \int \frac(x+2)(x^2+5x-6) dx = \frac(3)(7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$ |
“A mathematician, just like an artist or poet, creates patterns. And if his patterns are more stable, it is only because they are composed of ideas... The patterns of a mathematician, just like the patterns of an artist or a poet, must be beautiful; Ideas, just like colors or words, must correspond to each other. Beauty is the first requirement: there is no place in the world for ugly mathematics».
G.H.Hardy
In the first chapter it was noted that there are antiderivatives of fairly simple functions that can no longer be expressed through elementary functions. In this regard, those classes of functions about which we can accurately say that their antiderivatives are elementary functions acquire enormous practical importance. This class of functions includes rational functions, representing the ratio of two algebraic polynomials. Many problems lead to the integration of rational fractions. Therefore, it is very important to be able to integrate such functions.
2.1.1. Fractional rational functions
Rational fraction(or fractional rational function) is called the relation of two algebraic polynomials:
where and are polynomials.
Let us recall that polynomial (polynomial, whole rational function) nth degree called a function of the form
Where 
– real numbers. For example,
– polynomial of the first degree;
– polynomial of the fourth degree, etc.
The rational fraction (2.1.1) is called correct, if the degree is lower than the degree , i.e. n<m, otherwise the fraction is called wrong.
Any improper fraction can be represented as the sum of a polynomial (integer part) and a proper fraction (fractional part). The separation of the whole and fractional parts of an improper fraction can be done according to the rule for dividing polynomials with a “corner”.
Example 2.1.1. Identify the whole and fractional parts of the following improper rational fractions:
A) 
, b) 
.
Solution . a) Using the “corner” division algorithm, we get
Thus, we get
.
b) Here we also use the “corner” division algorithm:
As a result, we get
.
Let's summarize. In the general case, the indefinite integral of a rational fraction can be represented as the sum of the integrals of the polynomial and the proper rational fraction. Finding antiderivatives of polynomials is not difficult. Therefore, in what follows we will mainly consider proper rational fractions.
2.1.2. The simplest rational fractions and their integration
Among proper rational fractions, there are four types, which are classified as the simplest (elementary) rational fractions:
|
3) |
4) |
,
,where is an integer, 
, i.e. quadratic trinomial 
has no real roots.
Integrating simple fractions of type 1 and type 2 does not present much difficulty:
, (2.1.3)
. (2.1.4)
Let us now consider the integration of simple fractions of the 3rd type, but we will not consider fractions of the 4th type.
Let's start with integrals of the form
.
This integral is usually calculated by isolating the perfect square of the denominator. The result is a table integral of the following form
or 
.
Example 2.1.2. Find the integrals:
A) 
, b) 
.
Solution . a) Select a complete square from a quadratic trinomial:
From here we find
b) By isolating a complete square from a quadratic trinomial, we obtain:
Thus,
.
To find the integral
you can isolate the derivative of the denominator in the numerator and expand the integral into the sum of two integrals: the first of them by substitution 
comes down to appearance
,
and the second - to the one discussed above.
Example 2.1.3. Find the integrals:
.
Solution
. Note that 
. Let us isolate the derivative of the denominator in the numerator:
The first integral is calculated using substitution 
:
In the second integral, we select the perfect square in the denominator
Finally, we get
2.1.3. Proper rational fraction expansion
for the sum of simple fractions
Any proper rational fraction 
can be represented in a unique way as a sum of simple fractions. To do this, the denominator must be factorized. From higher algebra it is known that every polynomial with real coefficients
To integrate a rational function \(\large\frac((P\left(x \right)))((Q\left(x \right)))\normalsize,\) where \((P\left(x \right ))\) and \((Q\left(x \right))\) are polynomials, the following sequence of steps is used:
If the fraction is an improper fraction (i.e. the degree of \((P\left(x \right))\) is greater than the degree of \((Q\left(x \right))\)), convert it to a proper fraction by highlighting the whole expression;
Expand the denominator \((Q\left(x \right))\) into the product of monomials and/or irreducible quadratic expressions;
Resolve a rational fraction into simpler fractions using ;
Calculate integrals of simple fractions.
Step 1: Converting an improper rational fraction
If the fraction is improper (i.e. the degree of the numerator \((P\left(x \right))\) is greater than the degree of the denominator \((Q\left(x \right))\)), divide the polynomial \((P\ left(x \right))\) by \((Q\left(x \right)).\) We obtain the following expression: \[\frac((P\left(x \right)))((Q\left (x \right))) = F\left(x \right) + \frac((R\left(x \right)))((Q\left(x \right))),\] where \(\ large\frac((R\left(x \right)))((Q\left(x \right)))\normalsize\) is a proper rational fraction.
Step 2. Decomposition of the denominator into simple fractions
Let us write the denominator polynomial \((Q\left(x \right))\) in the form \[ (Q\left(x \right) ) = ((\left((x - a) \right)^\alpha ) \ cdots (\left((x - b) \right)^\beta )(\left(((x^2) + px + q) \right)^\mu ) \cdots (\left(((x^2 ) + rx + s) \right)^\nu ),) \] where quadratic functions are irreducible, that is, they do not have real roots.
Step 3. Decomposition of a rational fraction into a sum of simple fractions.
Let us write the rational function in the following form: \[ (\frac((R\left(x \right)))((Q\left(x \right))) = \frac(A)((((\left(( x - a) \right))^\alpha ))) + \frac(((A_1)))((((\left((x - a) \right))^(\alpha - 1)))) + \ldots )\kern0pt (+ \frac(((A_(\alpha - 1))))((x - a)) + \ldots )\kern0pt (+ \frac(B)((((\left( (x - b) \right))^\beta ))) + \frac(((B_1)))((((\left((x - b) \right))^(\beta - 1))) ) + \ldots )\kern0pt (+ \frac(((B_(\beta - 1))))((x - b)) )\kern0pt (+ \frac((Kx + L))((((\ left(((x^2) + px + q) \right))^\mu ))) + \frac(((K_1)x + (L_1)))((((\left(((x^2 ) + px + q) \right))^(\mu - 1)))) + \ldots )\kern0pt (+ \frac(((K_(\mu - 1))x + (L_(\mu - 1 ))))(((x^2) + px + q)) + \ldots )\kern0pt (+ \frac((Mx + N))((((\left(((x^2) + rx + s) \right))^\nu ))) + \frac(((M_1)x + (N_1)))((((\left(((x^2) + rx + s) \right))^ (\nu - 1)))) + \ldots )\kern0pt (+ \frac(((M_(\nu - 1))x + (N_(\nu - 1))))(((x^2) + rx + s)).) \] Total number of uncertain coefficients \((A_i),\) \((B_i),\) \((K_i),\) \((L_i),\) \((M_i ),\) \((N_i), \ldots\) must be equal to the degree of the denominator \((Q\left(x \right)).\)
Then we multiply both sides of the resulting equation by the denominator \((Q\left(x \right))\) and equate the coefficients of terms with the same degrees \(x.\) As a result, we obtain a system of linear equations for unknown coefficients \((A_i ),\) \((B_i),\) \((K_i),\) \((L_i),\) \((M_i),\) \((N_i), \ldots\) This system always has the only solution. The described algorithm is method of uncertain coefficients .
Step 4. Integration of simple rational fractions.
The simplest fractions obtained by decomposing an arbitrary proper rational fraction are integrated using the following six formulas: \ \ For fractions with a quadratic denominator, you first need to isolate the perfect square: \[\int (\frac((Ax + B))((((\ left(((x^2) + px + q) \right))^k)))dx) = \int (\frac((At + B"))((((\left(((t^2 ) + (m^2)) \right))^k)))dt) ,\] where \(t = x + \large\frac(p)(2)\normalsize,\) \((m^2 ) = \large\frac((4q - (p^2)))(4)\normalsize,\) \(B" = B - \large\frac((Ap))(2)\normalsize.\) Then the following formulas are used: \ \[ (4.\;\;\int (\frac((tdt))((((\left(((t^2) + (m^2)) \right))^k )))) ) = (\frac(1)((2\left((1 - k) \right)((\left(((t^2) + (m^2)) \right))^( k - 1)))) ) \] \Integral \(\large\int\normalsize (\large\frac((dt))((((\left(((t^2) + (m^2)) \right))^k)))\normalsize) \) can be calculated in \(k\) steps using reduction formulas\[ (6.\;\;\int (\frac((dt))((((\left(((t^2) + (m^2)) \right))^k)))) ) = (\frac(t)((2(m^2)\left((k - 1) \right)((\left(((t^2) + (m^2)) \right))^( k - 1)))) ) (+ \frac((2k - 3))((2(m^2)\left((k - 1) \right)))\int (\frac((dt)) ((((\left(((t^2) + (m^2)) \right))^(k - 1))))) ) \]
The problem of finding the indefinite integral of a fractionally rational function comes down to integrating simple fractions. Therefore, we recommend that you first familiarize yourself with the section of the theory of decomposition of fractions into the simplest.
Example.
Find the indefinite integral.
Solution.
Since the degree of the numerator of the integrand is equal to the degree of the denominator, we first select the whole part by dividing the polynomial by the polynomial with a column: 
That's why, 
.
The decomposition of the resulting proper rational fraction into simpler fractions has the form 
. Hence, 
The resulting integral is the integral of the simplest fraction of the third type. Looking ahead a little, we note that you can take it by subsuming it under the differential sign.
Because 
, That 
. That's why 
Hence, 
Now let's move on to describing methods for integrating simple fractions of each of the four types.
Integration of simple fractions of the first type
The direct integration method 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 simple fractions of the second type
The direct integration method is also suitable for solving this problem: 
Example.
Solution.
Top of page
Integration of simple fractions of the third type 
First we present the indefinite integral 
as a sum:
We take the first integral by subsuming it under the differential sign: 
That's why, 
Let us transform the denominator of the resulting integral: 
Hence, 
The formula for integrating simple 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 simple fractions of the fourth type 
The first step is to put it under the differential sign: 
The second step is to find an integral of the form 
. Integrals of this type are found using recurrence formulas. (See section on integration using recurrence formulas.) The following recurrent formula is suitable for our case:
Example.
Find the indefinite integral
Solution.
For this type of integrand we use the substitution method. Let's introduce a new variable (see the section on integration of irrational functions): 
After substitution we have: 
We came to finding the integral of a fraction of the fourth type. In our case we have coefficients M = 0, p = 0, q = 1, N = 1 And n=3. We apply the recurrent formula: 
After reverse replacement we get the result:
| Integrating trigonometric functions | ||||||||||||||||||||
1.Integrals of the form    are calculated by transforming the product of trigonometric functions into a sum using the formulas:  For example, 2.Integrals of the form  , Where m or n– an odd positive number, calculated by subsuming it under the differential sign. For example,   3.Integrals of the form , Where m And n–even positive numbers are calculated using formulas for reducing the degree: 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 a universal trigonometric substitution then (since  =[after dividing the numerator and denominator by ]= ;  For example,   It should be noted that the use of universal substitution often leads to cumbersome calculations. |
||||||||||||||||||||
| §5. Integration of the simplest irrationalities | ||||||||||||||||||||
Let's consider methods for integrating the simplest types of irrationality. 1.  Functions of this type are integrated in the same way as the simplest rational fractions of the 3rd type: in the denominator, a complete square is isolated from the square trinomial and a new variable is introduced. Example. 
2.  (under the integral sign – rational function of arguments). Integrals of this type are calculated using substitution. In particular, in integrals of the form we denote . If the integrand contains roots of different degrees:  , then denote where n– least common multiple of numbers m,k. Example 1.  Example 2.  -improper rational fraction, select the whole part:
3.Integrals of the form
|
are calculated using trigonometric substitutions:
44 
45 Definite integral
Definite integral- 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 a domain in the set of specifying this function (functional).
Definition
Let it be defined on . Let's divide it into parts with several arbitrary points. Then they say that the segment has been partitioned. Next, choose an arbitrary point 
, ,
A definite integral of a function on an interval is the limit of integral sums as the rank of the partition tends to zero, if it exists independently of the partition and choice of points, that is
If the specified limit exists, then the function is said to be Riemann integrable.
Designations
· - lower limit.
· - upper limit.
· - integrand function.
· - length of the partial segment.
· - 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 meaning
Definite integral as the area of a figure
The definite integral is numerically equal to the area of the figure bounded by the abscissa axis, straight lines and the graph of the function.
Newton-Leibniz theorem
[edit]
(redirected from "Newton-Leibniz Formula")
Newton-Leibniz formula or main theorem of analysis gives the relationship between two operations: taking a definite integral and calculating the antiderivative.
Proof
Let an integrable function be given on an interval. Let's start by noting that
that is, it does not matter which letter (or) is under the sign in the definite integral over the segment.
Let's 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 . Let us recall that we consider by definition
(1)
Note that
Let us show that it is continuous on the interval . In fact, let ; Then
and if , then
Thus, it is continuous regardless of whether it has or does not have discontinuities; 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 its boundedness, obviously tends to zero at, regardless of whether it is a point of continuity or discontinuity, for example a point.
Let now the function not only be integrable on , but continuous at the point . Let us prove that then the derivative at this point is equal to
(2)
In fact, for the indicated point
(1) , (3)
We put , and since it is constant relative to ,TO 
. Further, due to continuity at a point, for any one can specify such that for .
which proves that the left-hand 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). When we are talking here about the right and left derivatives, respectively.
If a function is continuous on , then, based on what was proven above, the corresponding function
(4)
has a derivative equal to . Therefore, the function is an antiderivative for .
This conclusion is sometimes called the variable upper bound integral theorem or Barrow's theorem.
We have proven that an arbitrary function continuous 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.
Let now there be an arbitrary antiderivative of a function on . We know that , where is some constant. Assuming in this equality and taking into account that , we obtain .
Thus, . But
Improper integral
[edit]
Material from Wikipedia - the free encyclopedia
Definite integral called not your own, if at least one of the following conditions is met:
· Limit a or b (or both limits) are 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 is used 
and the integral is called improper Riemann integral of the first kind. In this case is called convergent.
2. If there is no finite 
( or ), then the integral is said to diverge to , or simply divergent.
If a function is defined and continuous on the entire number line, then there may be an improper integral of this function with two infinite limits of integration, defined by the formula:
, where c is an arbitrary number.
[edit] Geometric meaning of an improper integral of the first kind
The improper integral expresses the area of an infinitely long curved 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 is used 
and the integral is called
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 is used 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 remains the same, and called divergent to , or simply divergent.
If the function suffers a discontinuity at an interior 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 tall curved trapezoid
[edit] Example
[edit]Isolated case
Let the function be defined on the entire number line and have a discontinuity at the points.
Then we can find the improper integral
[edit] Cauchy criterion
1. Let it be defined on a set from and 
.
Then 
converges
2. Let be defined on and 
.
Then converges
[edit]Absolute convergence
Integral 
called absolutely convergent, If 
converges.
If the integral converges absolutely, then it converges.
[edit]Conditional convergence
The integral is called conditionally convergent, if it converges, but diverges.
48 12. Improper integrals.
When considering definite integrals, we assumed that the domain of integration is limited (more specifically, it is the segment [ a ,b ]); For the existence of a definite integral, the integrand must be bounded on [ a ,b ]. We will call definite integrals for which both of these conditions are satisfied (boundedness of both the domain of integration and the integrand) own; integrals for which these requirements are violated (i.e., either the integrand or the domain of integration is unlimited, or both together) not your 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. Newton-Leibniz formula for an 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 its extreme form.
- 12.1.4. Absolute convergence of improper integrals over an infinite interval.
- 12.1.5. Tests for Abel and Dirichlet convergence.
- 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. The singularity is at the left end of the integration interval.
- 12.2.1.2. Application of the Newton-Leibniz formula.
- 12.2.1.3. The singularity at the right end of the integration interval.
- 12.2.1.4. Singularity at the inner point of the integration interval.
- 12.2.1.5. Several features on the integration interval.
- 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 its extreme form.
- 12.2.3. Absolute and conditional convergence of improper integrals of discontinuous functions.
- 12.2.4. Tests for Abel and Dirichlet convergence.
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 semi-axis and is integrable over any interval [ from, implying in each of these cases the existence and finitude of the corresponding limits. Now the solutions to the examples look simpler: 
.
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 tendency ( or ), then the integral converges, otherwise it diverges. When solving practical problems, however, it is important to first establish the fact of convergence itself, and only then calculate the integral (besides, the antiderivative is often not expressed in terms of elementary functions). Let us 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
) integral
The derivation of formulas for calculating integrals of 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 integrating a fraction of the fourth type is considered.
ContentSee 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 the simplest, 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 has no real roots.
Integration of fractions of the first two types
Integrating the first two fractions is done using the following formulas from the table of integrals:
,
, n ≠ - 1
.
1. Integration of fractions of the first type
A fraction of the first type is reduced to a table integral by substitution t = x - a:
.
2. Integration of fractions of the second type
The fraction of the second type is reduced to a table integral by the same substitution t = x - a:
.
3. Integration of fractions 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
Let us isolate the derivative of the denominator in the numerator of the fraction. Let us denote: u = x 2 + bx + c. Let's differentiate: u′ = 2 x + b. Then
;
.
But
.
We omitted the modulus 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 found the integral of a fraction of the third type:
,
Where .
4. Integration of fractions of the fourth type
And finally, consider the integral of a fraction of the fourth type:
.
We calculate it in three steps.
4.1) Select the derivative of the denominator in the numerator:
.
4.2) Calculate the integral
.
4.3) Calculate integrals
,
using the reduction formula:
.
4.1. Step 1. Isolating the derivative of the denominator in the numerator
Let us isolate the derivative of the denominator in the numerator, as we did in . Let us denote u = x 2 + bx + c. Let's differentiate: u′ = 2 x + b. Then
.
.
But
.
Finally we have:
.
4.2. Step 2. Calculate the integral with n = 1
Calculate the integral
.
Its calculation is outlined in .
4.3. Step 3. Derivation of the reduction formula
Now consider the integral
.
We reduce the quadratic trinomial to the sum of squares:
.
Here .
Let's make a substitution.
.
.
We carry out transformations and integrate in parts.
.
Multiply by 2(n - 1):
.
Let's return to x and I n.
,
;
;
.
So, for I n we got the reduction formula:
.
Consistently applying this formula, we reduce the integral I n to I 1
.
Example
Calculate integral
1.
Let us isolate 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:
.
Find the coefficient for .
.
