“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 primitives quite simple functions, which 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 remind you 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 (the whole part) and a proper fraction (the 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 the 1st and 2nd types does not present any great difficulties:
, (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 full square in 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
. notice, that 
. Let us isolate the derivative of the denominator in the numerator:
The first integral is calculated using the 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
A rational function is a fraction of the form , the numerator and denominator of which are polynomials or products of polynomials.
Example 1. Step 2.
.
We multiply the undetermined coefficients by polynomials that are not in this individual fraction, but which are in other resulting fractions:
We open the brackets and equate the numerator of the original integrand to the resulting expression:
In both sides of the equality, we look for terms with the same powers of x and compose a system of equations from them:
.
We cancel all the x’s and get an equivalent system of equations:
.
Thus, the final expansion of the integrand into a sum of simple fractions is:
.
Example 2. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
.
Now we begin to look for uncertain coefficients. To do this, we equate the numerator of the original fraction in the function expression to the numerator of the expression obtained after reducing the sum of fractions to a common denominator:
Now you need to create and solve a system of equations. To do this, we equate the coefficients of the variable to the corresponding degree in the numerator of the original expression of the function and similar coefficients in the expression obtained at the previous step:
We solve the resulting system:
So, from here
.
Example 3. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
We begin to look for uncertain coefficients. To do this, we equate the numerator of the original fraction in the function expression to the numerator of the expression obtained after reducing the sum of fractions to a common denominator:
As in previous examples, we compose a system of equations:
We reduce the x's and get an equivalent system of equations:
Solving the system, we obtain the following values of the uncertain coefficients:
We obtain the final decomposition of the integrand into the sum of simple fractions:
.
Example 4. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
.
We already know from previous examples how to equate the numerator of the original fraction with the expression in the numerator obtained after decomposing the fraction into the sum of simple fractions and bringing this sum to a common denominator. Therefore, just for control purposes, we present the resulting system of equations:
Solving the system, we obtain the following values of the uncertain coefficients:
We obtain the final decomposition of the integrand into the sum of simple fractions:
Example 5. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
.
We independently reduce this sum to a common denominator, equating the numerator of this expression to the numerator of the original fraction. The result should be the following system of equations:
Solving the system, we obtain the following values of the uncertain coefficients:
.
We obtain the final decomposition of the integrand into the sum of simple fractions:
.
Example 6. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
We perform the same actions with this amount as in the previous examples. The result should be the following system of equations:
Solving the system, we obtain the following values of the uncertain coefficients:
.
We obtain the final decomposition of the integrand into the sum of simple fractions:
.
Example 7. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
.
After certain actions with the resulting amount, the following system of equations should be obtained:
Solving the system, we obtain the following values of the uncertain coefficients:
We obtain the final decomposition of the integrand into the sum of simple fractions:
.
Example 8. Step 2. At step 1, we obtained the following decomposition of the original fraction into the sum of simple fractions with undetermined coefficients in the numerators:
.
Let's make some changes to the actions that have already been brought to automaticity to obtain a system of equations. There is an artificial technique that in some cases helps to avoid unnecessary calculations. Bringing the sum of fractions to a common denominator, we obtain and equating the numerator of this expression to the numerator of the original fraction, we obtain.
TOPIC: Integration of rational fractions.
Attention! When studying one of the basic methods of integration: the integration of rational fractions, it is required to consider polynomials in the complex domain to carry out rigorous proofs. Therefore it is necessary study in advance some properties of complex numbers and operations on them.
Integration of simple rational fractions.
If P(z) And Q(z) are polynomials in the complex domain, then they are rational fractions. It is called correct, if degree P(z) less degree Q(z) , And wrong, if degree R no less than a 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 comes down 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 the following types of fractions:
1) , 2) , 3) , 4) 
.
Let's find out how they integrate.
3) 
(studied previously).
Theorem 5. Every 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 decomposition 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 among the roots of the polynomial there are only multiple real roots, then in the decomposition 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 decomposition 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 decomposition 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 given expansions proceed as follows. The left and right sides of the expansion containing unknown coefficients are multiplied by The equality of two polynomials is obtained. From it, equations for the required coefficients are obtained using:
1. equality is true for any values of X (partial value method). In this case, any number of equations are obtained, any m of which allow one to find the unknown coefficients.
2. the coefficients coincide for the same degrees of X (method uncertain coefficients). In this case, a system of m - equations with m - unknowns is obtained, from which the unknown coefficients are found.
3. combined method.
Example 5. Expand a fraction 
to the simplest.
Solution:
Let's find the coefficients A and B.
Method 1 - private value method:
Method 2 – method of undetermined coefficients:
Answer: 
Integrating 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 through elementary functions, namely rational fractions, logarithms and arctangents.
Proof.
Let's imagine a rational fraction in the form: 
. In this case, the last term is a proper fraction, and according to Theorem 5 it can be represented as a linear combination of simple fractions. Thus, the integration of a rational fraction is reduced to the integration of a polynomial S(x)
and simple fractions, the antiderivatives of which, as has been 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
All of the above in the previous paragraphs allows us to formulate the basic rules for the integration of rational fractions.
1. If a rational fraction is improper, then it is represented as the sum of a polynomial and a proper rational fraction (see paragraph 2).
This reduces the integration of an improper rational fraction to the integration of a polynomial and a proper rational fraction.
2. Factor the denominator of the proper fraction.
3. A proper rational fraction is decomposed into the sum of simple fractions. This reduces the integration of a proper rational fraction to the integration of simple fractions.
Let's look at examples.
Example 1. Find .
Solution. Below the integral is an improper rational fraction. Selecting the whole part, we get
Hence,
Noting that , let us expand the proper rational fraction
to simple fractions:
(see formula (18)). That's why
Thus, we finally have
Example 2. Find
Solution. Below the integral is a proper rational fraction.
Expanding it into simple fractions (see formula (16)), we obtain
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 highly recommend that you at least skim through this topic before moving on 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 corresponding 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 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 more complete understanding of the text): show\hide
Why is the condition $p^2-4q needed?< 0$ 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 at all necessary that the coefficient before $x^2$ be equal to 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 (proper and improper), as well as examples of decomposition of a rational fraction into elementary ones can be found. Here we will be interested only in questions of their integration. Let's start with the integration of elementary fractions. So, each of the four types of elementary fractions above is easy to integrate using the formulas below. Let me remind you that when integrating fractions of types (2) and (4), $n=2,3,4,\ldots$ are assumed. Formulas (3) and (4) require the fulfillment of 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 substitution $t=x+\frac(p)(2)$ is made, after which the resulting interval is divided into two. The first will be calculated by entering under the differential sign, and the second will have the form $I_n=\int\frac(dt)((t^2+a^2)^n)$. 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 discussed in example No. 7 (see the third part).
Scheme for calculating integrals of 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 you can integrate any rational fraction. That is why almost all changes of variables in an indefinite integral (Euler, Chebyshev, universal trigonometric substitution) are made in such a way that after this change we obtain a rational fraction under the interval. And then 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 $dx=d(x+9)$, 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 looking 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 it “manually”.
2) Again, there are two ways: use the ready-made formula or do without it. If you apply the formula, then you should take into account that the coefficient in front of $x$ (number 4) will have to be removed. To do this, let’s simply take this four out of 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 taking the constant $4$ out of brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, 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 for finding such integrals are given in the topic “Integration by substitution (substitution 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 really an elementary fraction of the third type, you need to check that the condition $p^2-4q is met< 0$. 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 a 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 will not last long. Let's apply the following transformation to the numerator:
$$ 4x+7=2\cdot 2x+7=2\cdot (2x+10-10)+7=2\cdot(2x+10)-2\cdot 10+7=2\cdot(2x+10) -13. $$
Now the required expression $2x+10$ appears 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 split the integrand into two. Well, and, accordingly, the integral itself is also “bifurcated”:
$$ \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx=\int \left(\frac(2\cdot(2x+10))(x^2 +10x+34)-\frac(13)(x^2+10x+34) \right)\; dx=\\ =\int \frac(2\cdot(2x+10))(x^2+10x+34)dx-\int\frac(13dx)(x^2+10x+34)=2\cdot \int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34). $$
Let's first talk about the first integral, 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 numerator of the integrand contains the differential of the denominator. 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 select a complete 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 we obtained earlier can be rewritten in a slightly different form:
$$ 2\cdot\int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34) =2\cdot \int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+5)^2+ 9). $$
If we make the substitution $u=x^2+10x+34$ in the first integral, then it will take the form $\int\frac(du)(u)$ and take easy to use second formula from . As for the second integral, the change $u=x+5$ is feasible for it, after which it will take the form $\int\frac(du)(u^2+9)$. This pure water eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we 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 received the same answer as when applying the formula, which, strictly speaking, is not surprising. In general, the formula is proved by the same methods that we used to find this integral. I believe that the attentive reader may have one question here, so I will formulate it:
Question No. 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 there no module in the solution?
Answer to question #1
The question is completely natural. The module was missing 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 $(x+5)^2+9 > 0$ . You can think differently, without using the selection of a complete square. Since $10^2-4\cdot 34=-16< 0$, то $x^2+10x+34 >0$ for any $x\in R$ (if this logical chain It's amazing, I recommend watching it graphic method solutions to quadratic inequalities). In any case, since $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. Instead of a module, you can use regular brackets.
All points of example No. 1 have been solved, all that remains is 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 No. 2
Find the integral $\int\frac(7x+12)(3x^2-5x-2)dx$.
At first glance, the integrand fraction $\frac(7x+12)(3x^2-5x-2)$ is very similar to an elementary fraction of the third type, i.e. by $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient of $3$ in front of $x^2$, but it doesn’t take long to remove the coefficient (put it out of brackets). However, this similarity is apparent. For the fraction $\frac(Mx+N)(x^2+px+q)$ the condition $p^2-4q is mandatory< 0$, 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 before $x^2$ is not equal to one, therefore 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 quadratic equation$x^2+px+q=0$. If the discriminant is less than zero, then the expression $x^2+px+q$ cannot be factorized. Let's calculate the discriminant of the polynomial $3x^2-5x-2$ located in the denominator of our fraction: $D=(-5)^2-4\cdot 3\cdot(-2)=49$. So, $D > 0$, therefore the expression $3x^2-5x-2$ can be factorized. This means that the fraction $\frac(7x+12)(3x^2-5x-2)$ is not an elemental fraction of the third type, and apply $\int\frac(7x+12)(3x^2-) to the integral 5x-2)dx$ formula is not possible.
Well, if the given rational fraction is not an elementary fraction, then it needs to be represented as a sum of elementary fractions and then integrated. In short, take advantage of the trail. 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 present the subintercal fraction in this 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 decompose 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 undetermined coefficients and the method of substitution of partial values. Let's apply 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, all that remains is 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 this entry, but I like a more accurate option:
$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=-\frac(29)(21)\ cdot\frac(1)(x+\frac(1)(3))+\frac(26)(7)\cdot\frac(1)(x-2). $$
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 place 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 No. 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 contains a polynomial of the second degree, and the denominator contains 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:
$$ \frac(x^2-38x+157)((x-1)(x+4)(x-9))=-\frac(3)(x-1)+\frac(5)(x +4)-\frac(1)(x-9). $$
All we have to do is split the given integral into three and apply the formula to each. I prefer to immediately place 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: $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=-3\ln|x-1|+5\ln|x+ 4|-\ln|x-9|+C$.
Continuation of the analysis of examples of this topic is located in the second part.
