## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}-x^{2}-7 x-12}{x(x-3)(x+1)} d x
$$
## Solution

$$
\int \frac{2 x^{3}-x^{2}-7 x-12}{x(x-3)(x+1)} d x=\int \frac{2 x^{3}-x^{2}-7 x-12}{x^{3}-2 x^{2}-3 x} d x=
$$

The integrand is an improper fraction. Let's separate the integer part:

$$
\begin{aligned}
& 2 x^{3}-x^{2}-7 x-12 \\
& \frac{2 x^{3}-4 x^{2}-6 x}{3 x^{2}-x-12}
\end{aligned}
$$

$$
\frac{\mid x^{3}-2 x^{2}-3 x}{2}
$$

We get:

$$
=\int\left(2+\frac{3 x^{2}-x-12}{x^{3}-2 x^{2}-3 x}\right) d x=\int 2 d x+\int \frac{3 x^{2}-x-12}{x(x-3)(x+1)} d x=
$$

Let's decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{3 x^{2}-x-12}{x(x-3)(x+1)}=\frac{A}{x}+\frac{B}{x-3}+\frac{C}{x+1}= \\
& =\frac{A(x-3)(x+1)+B x(x+1)+C x(x-3)}{x(x-3)(x+1)}= \\
& =\frac{A\left(x^{2}-2 x-3\right)+B\left(x^{2}+x\right)+C\left(x^{2}-3 x\right)}{x(x-3)(x+1)}= \\
& =\frac{(A+B+C) x^{2}+(-2 A+B-3 C) x-3 A}{x(x-3)(x+1)}= \\
& \left\{\begin{array}{l}
A+B+C=3 \\
-2 A+B-3 C=-1 \\
-3 A=-12
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=-1 \\
B-3 C=7 \\
A=4
\end{array}\right.
$$

Add the first row multiplied by 3 to the second row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=-1 \\
4 B=4 \\
A=4
\end{array}\right. \\
& \left\{\begin{array}{l}
C=-2 \\
B=1 \\
A=4
\end{array}\right. \\
& \frac{3 x^{2}-x-12}{x(x-3)(x+1)}=\frac{4}{x}+\frac{1}{x-3}-\frac{2}{x+1}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int 2 d x+\int\left(\frac{4}{x}+\frac{1}{x-3}-\frac{2}{x+1}\right) d x= \\
& =2 x+4 \cdot \ln |x|+\ln |x-3|-2 \cdot \ln |x+1|+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-30$ »

Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

Ukrainian Banner Network

- Last modified: 12:10, 7 April 2009.
- Content is available under CC-BY-SA 3.0.