As I already noted, in integral calculus there is no convenient formula for integrating a fraction. And therefore, there is a sad trend: the more “fancy” the fraction, the more difficult it is to find the integral from it. In this regard, one has to resort to various tricks, which I will now discuss. Prepared readers can immediately use table of contents:
- The method of subsuming under the sign of the differential for simple fractions
Numerator Artificial Transformation Method
Example 1
By the way, the considered integral can also be solved by the change of variable method, denoting , but the solution will be much longer.
Example 2
Find the indefinite integral. Run a check.
This is a do-it-yourself example. It should be noted that here the variable replacement method will no longer work.
Attention important! Examples No. 1, 2 are typical and are common. In particular, such integrals often arise in the course of solving other integrals, in particular, when integrating irrational functions (roots).
The above method also works in the case if the highest power of the numerator is greater than the highest power of the denominator.
Example 3
Find the indefinite integral. Run a check.
We begin to select the numerator.
The numerator selection algorithm is something like this:
1) In the numerator I need to organize , but there . What to do? I enclose in brackets and multiply by: .
2) Now I try to open these brackets, what happens? . Hmm ... already better, but there is no deuce with initially in the numerator. What to do? You need to multiply by:
3) Opening the brackets again: . And here is the first success! Needed turned out! But the problem is that an extra term has appeared. What to do? In order for the expression not to change, I must add the same to my construction: 
. Life has become easier. Is it possible to organize again in the numerator?
4) You can. We try: 
. Expand the brackets of the second term: 
. Sorry, but I actually had in the previous step, and not . What to do? We need to multiply the second term by: 
5) Again, for verification, I open the brackets in the second term: 
. Now it's normal: obtained from the final construction of paragraph 3! But again there is a small “but”, an extra term has appeared, which means that I must add to my expression: 
If everything is done correctly, then when opening all the brackets, we should get the original numerator of the integrand. We check:
Good.
Thus: 
Ready. In the last term, I applied the method of bringing the function under the differential.
If we find the derivative of the answer and bring the expression to a common denominator, then we get exactly the original integrand. The considered method of expansion into a sum is nothing more than the reverse action to bring the expression to a common denominator.
The numerator selection algorithm in such examples is best performed on a draft. With some skills, it will also work mentally. I remember a record time when I did a selection for the 11th power, and the expansion of the numerator took almost two lines of Werd.
Example 4
Find the indefinite integral. Run a check.
This is a do-it-yourself example.
The method of subsuming under the sign of the differential for simple fractions
Let's move on to the next type of fractions.
, , , (the coefficients and are not equal to zero).
In fact, a couple of cases with arcsine and arctangent have already slipped in the lesson Variable change method in indefinite integral. Such examples are solved by bringing the function under the sign of the differential and then integrating using the table. Here are some more typical examples with a long and high logarithm:
Example 5
Example 6
Here it is advisable to pick up a table of integrals and follow what formulas and as transformation takes place. Note, how and why squares are highlighted in these examples. In particular, in Example 6, we first need to represent the denominator as 
, then bring under the sign of the differential. And you need to do all this in order to use the standard tabular formula 
.
But what to look at, try to solve examples No. 7,8 on your own, especially since they are quite short:
Example 7
Example 8
Find the indefinite integral:
If you can also check these examples, then great respect is your differentiation skills at their best.
Full square selection method
Integrals of the form , 
(coefficients and are not equal to zero) are solved full square selection method, which has already appeared in the lesson Geometric Plot Transformations.
In fact, such integrals reduce to one of the four table integrals that we have just considered. And this is achieved using the familiar abbreviated multiplication formulas:
Formulas are applied in this direction, that is, the idea of the method is to artificially organize expressions in the denominator or , and then convert them, respectively, to or .
Example 9
Find the indefinite integral
This is the simplest example where with the term - unit coefficient(and not some number or minus).
We look at the denominator, here the whole thing is clearly reduced to the case. Let's start converting the denominator:
Obviously, you need to add 4. And so that the expression does not change - the same four and subtract:
Now you can apply the formula:
After the conversion is finished ALWAYS it is desirable to perform a reverse move: everything is fine, there are no errors.
The clean design of the example in question should look something like this: 
Ready. Bringing a "free" complex function under the differential sign: , in principle, could be neglected
Example 10
Find the indefinite integral:
This is an example for self-solving, the answer is at the end of the lesson.
Example 11
Find the indefinite integral:
What to do when there is a minus in front? In this case, you need to take the minus out of brackets and arrange the terms in the order we need:. Constant("double" in this case) do not touch!
Now we add one in parentheses. Analyzing the expression, we come to the conclusion that we need one behind the bracket - add:
Here is the formula, apply:
ALWAYS we perform a check on the draft:
, which was to be verified.
The clean design of the example looks something like this: 
We complicate the task
Example 12
Find the indefinite integral: 
Here, with the term, it is no longer a single coefficient, but a “five”.
(1) If a constant is found at, then we immediately take it out of brackets.
(2) In general, it is always better to take this constant out of the integral, so that it does not get in the way.
(3) It is obvious that everything will be reduced to the formula . It is necessary to understand the term, namely, to get a "two"
(4) Yep, . So, we add to the expression, and subtract the same fraction.
(5) Now select a full square. In the general case, it is also necessary to calculate , but here we have a long logarithm formula 
, and the action does not make sense to perform, why - it will become clear a little lower.
(6) Actually, we can apply the formula 
, only instead of "x" we have, which does not negate the validity of the tabular integral. Strictly speaking, one step is missing - before integration, the function should have been brought under the differential sign: 
, but, as I have repeatedly noted, this is often neglected.
(7) In the answer under the root, it is desirable to open all the brackets back:
Complicated? This is not the most difficult in integral calculus. Although, the examples under consideration are not so much complicated as they require good calculation technique.
Example 13
Find the indefinite integral:
This is a do-it-yourself example. Answer at the end of the lesson.
There are integrals with roots in the denominator, which, with the help of a replacement, are reduced to integrals of the considered type, you can read about them in the article Complex integrals, but it is designed for highly prepared students.
Bringing the numerator under the sign of the differential
This is the final part of the lesson, however, integrals of this type are quite common! If fatigue has accumulated, maybe it is better to read tomorrow? ;)
The integrals that we will consider are similar to the integrals of the previous paragraph, they have the form: or 
(the coefficients , and are not equal to zero).
That is, we have a linear function in the numerator. How to solve such integrals?
The fraction is called correct if the highest power of the numerator is less than the highest power of the denominator. The integral of a proper 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 select a full square from it: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(mx+n)(x^2 \pm a ^2) $$
- Different real roots $ x_1 $ and $ x_2 $, then you need to expand the integral and find the indefinite coefficients $ 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 this 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 mind by dividing the polynomial from the numerator by the polynomial from the denominator. In this case, the formula for integrating a rational fraction is:
$$ \int \frac(P(x))(ax^2+bx+c)dx = \int Q(x) dx + \int \frac(mx+n)(ax^2+bx+c)dx $$
Solution examples
| Example 1 |
| Find the integral of a rational fraction: $$ \int \frac(dx)(x^2-10x+16) $$ |
| Decision |
|
The fraction is regular and the polynomial has only complex roots. Therefore, we select a full square: $$ \int \frac(dx)(x^2-10x+16) = \int \frac(dx)(x^2-2\cdot 5 x+ 5^2 - 9) = $$ We collapse the full square and sum 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 get: $$ = \frac(1)(2 \cdot 3) \ln \bigg | \frac(x-5 - 3)(x-5 + 3) \bigg | + C = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$ If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the 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 |
| Integrate rational fractions: $$ \int \frac(x+2)(x^2+5x-6) dx $$ |
| Decision |
|
Solve the quadratic equation: $$ x^2+5x-6 = 0 $$ $$ x_(12) = \frac(-5\pm \sqrt(25-4\cdot 1 \cdot (-6)))(2) = \frac(-5 \pm 7)(2) $$ Let's 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: $$ \frac(x+2)((x-1)(x+6)) = \frac(A)(x-1) + \frac(B)(x+6) = \frac(A(x -6)+B(x-1))((x-1)(x+6)) $$ 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 |
| $$ \int \frac(x+2)(x^2+5x-6) dx = \frac(3)(7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$ |
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.
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 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 has no 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. So .
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 bring the 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 .
.
The material presented in this topic is based on the information presented in the topic "Rational fractions. Decomposition of rational fractions into elementary (simple) fractions". I strongly advise you to at least skim through this topic before proceeding to reading this material. In addition, we will need a table of indefinite integrals.
Let me remind you of a couple of terms. They were discussed in the relevant topic, so here I will limit myself to a brief formulation.
The ratio of two polynomials $\frac(P_n(x))(Q_m(x))$ is called a rational function or a rational fraction. The rational fraction is called correct if $n< m$, т.е. если степень многочлена, стоящего в числителе, меньше степени многочлена, стоящего в знаменателе. В противном случае (если $n ≥ m$) дробь называется wrong.
Elementary (simplest) rational fractions are rational fractions of four types:
- $\frac(A)(x-a)$;
- $\frac(A)((x-a)^n)$ ($n=2,3,4, \ldots$);
- $\frac(Mx+N)(x^2+px+q)$ ($p^2-4q< 0$);
- $\frac(Mx+N)((x^2+px+q)^n)$ ($p^2-4q< 0$; $n=2,3,4,\ldots$).
Note (desirable for a better understanding of the text): show\hide
Why is the $p^2-4q condition necessary?< 0$ в дробях третьего и четвертого типов? Рассмотрим квадратное уравнение $x^2+px+q=0$. Дискриминант этого уравнения $D=p^2-4q$. По сути, условие $p^2-4q < 0$ означает, что $D < 0$. Если $D < 0$, то уравнение $x^2+px+q=0$ не имеет действительных корней. Т.е. выражение $x^2+px+q$ неразложимо на множители. Именно эта неразложимость нас и интересует.
For example, for the expression $x^2+5x+10$ we get: $p^2-4q=5^2-4\cdot 10=-15$. Since $p^2-4q=-15< 0$, то выражение $x^2+5x+10$ нельзя разложить на множители.
By the way, for this check it is not necessary that the coefficient in front of $x^2$ equals 1. For example, for $5x^2+7x-3=0$ we get: $D=7^2-4\cdot 5 \cdot (-3)=109$. Since $D > 0$, the expression $5x^2+7x-3$ is factorizable.
Examples of rational fractions (regular and improper), as well as examples of the expansion of a rational fraction into elementary ones, can be found. Here we are only interested in questions of their integration. Let's start with the integration of elementary fractions. So, each of the four types of the above elementary fractions is easy to integrate using the formulas below. Let me remind you that when integrating fractions of type (2) and (4) $n=2,3,4,\ldots$ is assumed. Formulas (3) and (4) require the condition $p^2-4q< 0$.
\begin(equation) \int \frac(A)(x-a) dx=A\cdot \ln |x-a|+C \end(equation) \begin(equation) \int\frac(A)((x-a)^n )dx=-\frac(A)((n-1)(x-a)^(n-1))+C \end(equation) \begin(equation) \int \frac(Mx+N)(x^2 +px+q) dx= \frac(M)(2)\cdot \ln (x^2+px+q)+\frac(2N-Mp)(\sqrt(4q-p^2))\arctg\ frac(2x+p)(\sqrt(4q-p^2))+C \end(equation)
For $\int\frac(Mx+N)((x^2+px+q)^n)dx$ the replacement $t=x+\frac(p)(2)$ is made, after which the resulting integral is split into two. The first one will be calculated by inserting it under the differential sign, and the second one will look like $I_n=\int\frac(dt)((t^2+a^2)^n)$. This integral is taken using the recurrence relation
\begin(equation) I_(n+1)=\frac(1)(2na^2)\frac(t)((t^2+a^2)^n)+\frac(2n-1)(2na ^2)I_n, \; n\in N \end(equation)
The calculation of such an integral is analyzed in example No. 7 (see the third part).
Scheme for calculating integrals from rational functions (rational fractions):
- If the integrand is elementary, then apply formulas (1)-(4).
- If the integrand is not elementary, then represent it as a sum of elementary fractions, and then integrate using formulas (1)-(4).
The above algorithm for integrating rational fractions has an undeniable advantage - it is universal. Those. Using this algorithm, one can integrate any rational fraction. That is why almost all replacements of variables in the indefinite integral (Euler, Chebyshev substitutions, universal trigonometric substitution) are done in such a way that after this replacement we get a rational fraction under the interval. And apply the algorithm to it. We will analyze the direct application of this algorithm using examples, after making a small note.
$$ \int\frac(7dx)(x+9)=7\ln|x+9|+C. $$
In principle, this integral is easy to obtain without mechanical application of the formula. If we take the constant $7$ out of the integral sign and take into account that $dx=d(x+9)$, then we get:
$$ \int\frac(7dx)(x+9)=7\cdot \int\frac(dx)(x+9)=7\cdot \int\frac(d(x+9))(x+9 )=|u=x+9|=7\cdot\int\frac(du)(u)=7\ln|u|+C=7\ln|x+9|+C. $$
For detailed information I recommend to look at the topic. It explains in detail how such integrals are solved. By the way, the formula is proved by the same transformations that were applied in this paragraph when solving "manually".
2) Again, there are two ways: to apply a ready-made formula or to do without it. If you apply the formula, then you should take into account that the coefficient in front of $x$ (the number 4) will have to be removed. To do this, we simply take out the four of them in brackets:
$$ \int\frac(11dx)((4x+19)^8)=\int\frac(11dx)(\left(4\left(x+\frac(19)(4)\right)\right)^ 8)= \int\frac(11dx)(4^8\left(x+\frac(19)(4)\right)^8)=\int\frac(\frac(11)(4^8)dx) (\left(x+\frac(19)(4)\right)^8). $$
Now it's time to apply the formula:
$$ \int\frac(\frac(11)(4^8)dx)(\left(x+\frac(19)(4)\right)^8)=-\frac(\frac(11)(4 ^8))((8-1)\left(x+\frac(19)(4) \right)^(8-1))+C= -\frac(\frac(11)(4^8)) (7\left(x+\frac(19)(4) \right)^7)+C=-\frac(11)(7\cdot 4^8 \left(x+\frac(19)(4) \right )^7)+C. $$
You can do without using the formula. And even without putting the constant $4$ out of the brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, then we get:
$$ \int\frac(11dx)((4x+19)^8)=11\int\frac(dx)((4x+19)^8)=\frac(11)(4)\int\frac( d(4x+19))((4x+19)^8)=|u=4x+19|=\\ =\frac(11)(4)\int\frac(du)(u^8)=\ frac(11)(4)\int u^(-8)\;du=\frac(11)(4)\cdot\frac(u^(-8+1))(-8+1)+C= \\ =\frac(11)(4)\cdot\frac(u^(-7))(-7)+C=-\frac(11)(28)\cdot\frac(1)(u^7 )+C=-\frac(11)(28(4x+19)^7)+C. $$
Detailed explanations on finding such integrals are given in the topic "Integration by substitution (introduction under the differential sign)" .
3) We need to integrate the fraction $\frac(4x+7)(x^2+10x+34)$. This fraction has the structure $\frac(Mx+N)(x^2+px+q)$, where $M=4$, $N=7$, $p=10$, $q=34$. However, to make sure that this is indeed an elementary fraction of the third type, you need to check the condition $p^2-4q< 0$. Так как $p^2-4q=10^2-4\cdot 34=-16 < 0$, то мы действительно имеем дело с интегрированием элементарной дроби третьего типа. Как и в предыдущих пунктах есть два пути для нахождения $\int\frac{4x+7}{x^2+10x+34}dx$. Первый путь - банально использовать формулу . Подставив в неё $M=4$, $N=7$, $p=10$, $q=34$ получим:
$$ \int\frac(4x+7)(x^2+10x+34)dx = \frac(4)(2)\cdot \ln (x^2+10x+34)+\frac(2\cdot 7-4\cdot 10)(\sqrt(4\cdot 34-10^2)) \arctg\frac(2x+10)(\sqrt(4\cdot 34-10^2))+C=\\ = 2\cdot \ln (x^2+10x+34)+\frac(-26)(\sqrt(36)) \arctg\frac(2x+10)(\sqrt(36))+C =2\cdot \ln (x^2+10x+34)+\frac(-26)(6) \arctg\frac(2x+10)(6)+C=\\ =2\cdot \ln (x^2+10x +34)-\frac(13)(3) \arctg\frac(x+5)(3)+C. $$
Let's solve the same example, but without using the ready-made formula. Let's try to isolate the derivative of the denominator in the numerator. What does this mean? We know that $(x^2+10x+34)"=2x+10$. It is the expression $2x+10$ that we have to isolate in the numerator. So far, the numerator contains only $4x+7$, but this is not for long. Apply the following transformation to the numerator:
$$ 4x+7=2\cdot 2x+7=2\cdot (2x+10-10)+7=2\cdot(2x+10)-2\cdot 10+7=2\cdot(2x+10) -thirteen. $$
Now the required expression $2x+10$ has appeared in the numerator. And our integral can be rewritten as follows:
$$ \int\frac(4x+7)(x^2+10x+34) dx= \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx. $$
Let's break the integrand into two. Well, and, accordingly, the integral itself is also "split":
$$ \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx=\int \left(\frac(2\cdot(2x+10))(x^2 +10x+34)-\frac(13)(x^2+10x+34) \right)\; dx=\\ =\int \frac(2\cdot(2x+10))(x^2+10x+34)dx-\int\frac(13dx)(x^2+10x+34)=2\cdot \int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34). $$
Let's talk about the first integral first, i.e. about $\int \frac((2x+10)dx)(x^2+10x+34)$. Since $d(x^2+10x+34)=(x^2+10x+34)"dx=(2x+10)dx$, then the denominator differential is located in the numerator of the integrand. In short, instead of the expression $( 2x+10)dx$ we write $d(x^2+10x+34)$.
Now let's say a few words about the second integral. Let's single out the full square in the denominator: $x^2+10x+34=(x+5)^2+9$. In addition, we take into account $dx=d(x+5)$. Now the sum of integrals obtained by us earlier can be rewritten in a slightly different form:
$$ 2\cdot\int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34) =2\cdot \int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+5)^2+ nine). $$
If we make the change $u=x^2+10x+34$ in the first integral, then it will take the form $\int\frac(du)(u)$ and is taken by simply applying the second formula from . As for the second integral, the replacement $u=x+5$ is feasible for it, after which it takes the form $\int\frac(du)(u^2+9)$. This is the purest water, the eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we will have:
$$ 2\cdot\int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+ 5)^2+9) =2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x+5)(3)+C. $$
We got the same answer as when applying the formula , which, in fact, is not surprising. In general, the formula is proved by the same methods that we used to find this integral. I believe that an attentive reader may have one question here, therefore I will formulate it:
Question #1
If we apply the second formula from the table of indefinite integrals to the integral $\int \frac(d(x^2+10x+34))(x^2+10x+34)$, then we get the following:
$$ \int \frac(d(x^2+10x+34))(x^2+10x+34)=|u=x^2+10x+34|=\int\frac(du)(u) =\ln|u|+C=\ln|x^2+10x+34|+C. $$
Why was the module missing from the solution?
Answer to question #1
The question is completely legitimate. The modulus was absent only because the expression $x^2+10x+34$ for any $x\in R$ is greater than zero. This is quite easy to show in several ways. For example, since $x^2+10x+34=(x+5)^2+9$ and $(x+5)^2 ≥ 0$, then $(x+5)^2+9 > 0$ . It is possible to judge in a different way, without involving the selection of a full square. Since $10^2-4\cdot 34=-16< 0$, то $x^2+10x+34 >0$ for any $x\in R$ (if this logical chain is surprising, I advise you to look at the graphical method for solving square inequalities). In any case, since $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. you can use normal brackets instead of a module.
All points of example No. 1 are solved, it remains only to write down the answer.
Answer:
- $\int\frac(7dx)(x+9)=7\ln|x+9|+C$;
- $\int\frac(11dx)((4x+19)^8)=-\frac(11)(28(4x+19)^7)+C$;
- $\int\frac(4x+7)(x^2+10x+34)dx=2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x +5)(3)+C$.
Example #2
Find the integral $\int\frac(7x+12)(3x^2-5x-2)dx$.
At first glance, the integrand $\frac(7x+12)(3x^2-5x-2)$ is very similar to an elementary fraction of the third type, i.e. to $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient $3$ in front of $x^2$, but it won't take long to remove the coefficient (out of brackets). However, this similarity is apparent. For the fraction $\frac(Mx+N)(x^2+px+q)$ the condition $p^2-4q< 0$, которое гарантирует, что знаменатель $x^2+px+q$ нельзя разложить на множители. Проверим, как обстоит дело с разложением на множители у знаменателя нашей дроби, т.е. у многочлена $3x^2-5x-2$.
Our coefficient in front of $x^2$ is not equal to one, so check the condition $p^2-4q< 0$ напрямую мы не можем. Однако тут нужно вспомнить, откуда взялось выражение $p^2-4q$. Это всего лишь дискриминант квадратного уравнения $x^2+px+q=0$. Если дискриминант меньше нуля, то выражение $x^2+px+q$ на множители не разложишь. Вычислим дискриминант многочлена $3x^2-5x-2$, расположенного в знаменателе нашей дроби: $D=(-5)^2-4\cdot 3\cdot(-2)=49$. Итак, $D >0$, so the expression $3x^2-5x-2$ can be factorized. And this means that the fraction $\frac(7x+12)(3x^2-5x-2)$ is not an elementary fraction of the third type, and apply to the integral $\int\frac(7x+12)(3x^2- 5x-2)dx$ formula is not allowed.
Well, if the given rational fraction is not elementary, then it must be represented as a sum of elementary fractions, and then integrated. In short, trail take advantage of . How to decompose a rational fraction into elementary ones is written in detail. Let's start by factoring the denominator:
$$ 3x^2-5x-2=0;\\ \begin(aligned) & D=(-5)^2-4\cdot 3\cdot(-2)=49;\\ & x_1=\frac( -(-5)-\sqrt(49))(2\cdot 3)=\frac(5-7)(6)=\frac(-2)(6)=-\frac(1)(3); \\ & x_2=\frac(-(-5)+\sqrt(49))(2\cdot 3)=\frac(5+7)(6)=\frac(12)(6)=2.\ \ \end(aligned)\\ 3x^2-5x-2=3\cdot\left(x-\left(-\frac(1)(3)\right)\right)\cdot (x-2)= 3\cdot\left(x+\frac(1)(3)\right)(x-2). $$
We represent the subinternal fraction in the following form:
$$ \frac(7x+12)(3x^2-5x-2)=\frac(7x+12)(3\cdot\left(x+\frac(1)(3)\right)(x-2) )=\frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2)). $$
Now let's expand the fraction $\frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))$ into elementary ones:
$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2)) =\frac(A)(x+\frac( 1)(3))+\frac(B)(x-2)=\frac(A(x-2)+B\left(x+\frac(1)(3)\right))(\left(x+ \frac(1)(3)\right)(x-2));\\ \frac(7)(3)x+4=A(x-2)+B\left(x+\frac(1)( 3)\right). $$
To find the coefficients $A$ and $B$ there are two standard ways: the method of indeterminate coefficients and the method of substitution of partial values. Let's use the partial value substitution method, substituting $x=2$ and then $x=-\frac(1)(3)$:
$$ \frac(7)(3)x+4=A(x-2)+B\left(x+\frac(1)(3)\right).\\ x=2;\; \frac(7)(3)\cdot 2+4=A(2-2)+B\left(2+\frac(1)(3)\right); \; \frac(26)(3)=\frac(7)(3)B;\; B=\frac(26)(7).\\ x=-\frac(1)(3);\; \frac(7)(3)\cdot \left(-\frac(1)(3) \right)+4=A\left(-\frac(1)(3)-2\right)+B\left (-\frac(1)(3)+\frac(1)(3)\right); \; \frac(29)(9)=-\frac(7)(3)A;\; A=-\frac(29\cdot 3)(9\cdot 7)=-\frac(29)(21).\\ $$
Since the coefficients have been found, it remains only to write down the finished expansion:
$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=\frac(-\frac(29)( 21))(x+\frac(1)(3))+\frac(\frac(26)(7))(x-2). $$
In principle, you can leave such a record, but I like a more accurate version:
$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=-\frac(29)(21)\ cdot\frac(1)(x+\frac(1)(3))+\frac(26)(7)\cdot\frac(1)(x-2). $$
Returning to the original integral, we substitute the resulting expansion into it. Then we divide the integral into two, and apply the formula to each. I prefer to immediately take out the constants outside the integral sign:
$$ \int\frac(7x+12)(3x^2-5x-2)dx =\int\left(-\frac(29)(21)\cdot\frac(1)(x+\frac(1) (3))+\frac(26)(7)\cdot\frac(1)(x-2)\right)dx=\\ =\int\left(-\frac(29)(21)\cdot\ frac(1)(x+\frac(1)(3))\right)dx+\int\left(\frac(26)(7)\cdot\frac(1)(x-2)\right)dx =- \frac(29)(21)\cdot\int\frac(dx)(x+\frac(1)(3))+\frac(26)(7)\cdot\int\frac(dx)(x-2 )dx=\\ =-\frac(29)(21)\cdot\ln\left|x+\frac(1)(3)\right|+\frac(26)(7)\cdot\ln|x- 2|+C. $$
Answer: $\int\frac(7x+12)(3x^2-5x-2)dx=-\frac(29)(21)\cdot\ln\left|x+\frac(1)(3)\right| +\frac(26)(7)\cdot\ln|x-2|+C$.
Example #3
Find the integral $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx$.
We need to integrate the fraction $\frac(x^2-38x+157)((x-1)(x+4)(x-9))$. The numerator is a polynomial of the second degree, and the denominator is a polynomial of the third degree. Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. $2< 3$, то подынтегральная дробь является правильной. Разложение этой дроби на элементарные (простейшие) было получено в примере №3 на странице, посвящённой разложению рациональных дробей на элементарные. Полученное разложение таково:
$$ \frac(x^2-38x+157)((x-1)(x+4)(x-9))=-\frac(3)(x-1)+\frac(5)(x +4)-\frac(1)(x-9). $$
We will only have to break the given integral into three, and apply the formula to each. I prefer to immediately take out the constants outside the integral sign:
$$ \int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=\int\left(-\frac(3)(x-1) +\frac(5)(x+4)-\frac(1)(x-9) \right)dx=\\=-3\cdot\int\frac(dx)(x-1)+ 5\cdot \int\frac(dx)(x+4)-\int\frac(dx)(x-9)=-3\ln|x-1|+5\ln|x+4|-\ln|x- 9|+C. $$
Answer: $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=-3\ln|x-1|+5\ln|x+ 4|-\ln|x-9|+C$.
A continuation of the analysis of examples of this topic is located in the second part.
Recall that fractionally rational are called functions of the form $$ f(x) = \frac(P_n(x))(Q_m(x)), $$ in the general case being the ratio of two polynomials %%P_n(x)%% and %%Q_m(x)% %.
If %%m > n \geq 0%%, then a rational fraction is called correct, otherwise it is incorrect. Using the polynomial division rule, an improper rational fraction can be represented as the sum of a polynomial %%P_(n - m)%% of degree %%n - m%% and some proper fraction, i.e. $$ \frac(P_n(x))(Q_m(x)) = P_(n-m)(x) + \frac(P_l(x))(Q_n(x)), $$ where the degree is %%l%% of the polynomial %%P_l(x)%% is less than the degree %%n%% of the polynomial %%Q_n(x)%%.
Thus, the indefinite integral of a rational function can be represented as the sum of indefinite integrals of a polynomial and of a proper rational fraction.
Integrals of simple rational fractions
There are four types of proper rational fractions, which are classified as the simplest rational fractions:
- %%\displaystyle \frac(A)(x - a)%%,
- %%\displaystyle \frac(A)((x - a)^k)%%,
- %%\displaystyle \frac(Ax + B)(x^2 + px + q)%%,
- %%\displaystyle \frac(Ax + B)((x^2 + px + q)^k)%%,
where %%k > 1%% is an integer and %%p^2 - 4q< 0%%, т.е. квадратные уравнения не имеют действительных корней.
Calculation of indefinite integrals from fractions of the first two types
Calculating indefinite integrals of fractions of the first two types is easy: $$ \begin(array)(ll) \int \frac(A)(x - a) \mathrm(d)x &= A\int \frac(\mathrm (d)(x - a))(x - a) = A \ln |x - a| + C, \\ \\ \int \frac(A)((x - a)^k) \mathrm(d)x &= A\int \frac(\mathrm(d)(x - a))(( x - a)^k) = A \frac((x-a)^(-k + 1))(-k + 1) + C = \\ &= -\frac(A)((k-1)(x-a )^(k-1)) + C. \end(array) $$
Calculation of indefinite integrals from fractions of the third type
We first transform the fraction of the third type by selecting the full square in the denominator: $$ \frac(Ax + B)(x^2 + px + q) = \frac(Ax + B)((x + p/2)^2 + q - p^2/4), $$ since %%p^2 - 4q< 0%%, то %%q - p^2/4 >0%%, which we will denote as %%a^2%%. Replacing also %%t = x + p/2, \mathrm(d)t = \mathrm(d)x%%, we transform the denominator and write the integral of a fraction of the third type in the form $$ \begin(array)(ll) \ int \frac(Ax + B)(x^2 + px + q) \mathrm(d)x &= \int \frac(Ax + B)((x + p/2)^2 + q - p^2 /4) \mathrm(d)x = \\ &= \int \frac(A(t - p/2) + B)(t^2 + a^2) \mathrm(d)t = \int \frac (At + (B - A p/2))(t^2 + a^2) \mathrm(d)t. \end(array) $$
Using the linearity of the indefinite integral, we represent the last integral as a sum of two and in the first of them we introduce %%t%% under the differential sign: $$ \begin(array)(ll) \int \frac(At + (B - A p /2))(t^2 + a^2) \mathrm(d)t &= A\int \frac(t \mathrm(d)t)(t^2 + a^2) + \left(B - \frac(pA)(2)\right)\int \frac(\mathrm(d)t)(t^2 + a^2) = \\ &= \frac(A)(2) \int \frac( \mathrm(d)\left(t^2 + a^2\right))(t^2 + a^2) + - \frac(2B - pA)(2)\int \frac(\mathrm(d) t)(t^2 + a^2) = \\ &= \frac(A)(2) \ln \left| t^2 + a^2\right| + \frac(2B - pA)(2a) \text(arctg)\frac(t)(a) + C. \end(array) $$
Returning to the original variable %%x%%, we end up with $$ \int \frac(Ax + B)(x^2 + px + q) \mathrm(d)x = \frac(A)( 2) \ln \left| x^2 + px + q\right| + \frac(2B - pA)(2a) \text(arctg)\frac(x + p/2)(a) + C, $$ where %%a^2 = q - p^2 / 4 > 0% %.
The calculation of the type 4 integral is difficult, so it is not covered in this course.
