## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{5}-x^{4}-6 x^{3}+13 x+6}{x(x-3)(x+2)} d x
$$
## Solution

$$
\int \frac{x^{5}-x^{4}-6 x^{3}+13 x+6}{x(x-3)(x+2)} d x=\int \frac{x^{5}-x^{4}-6 x^{3}+13 x+6}{x^{3}-x^{2}-6 x} d x=
$$

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

![](https://cdn.mathpix.com/cropped/2024_05_22_acc5b9fb81eba115622ag-53.jpg?height=202&width=910&top_left_y=1224&top_left_x=213)

We get:

$$
=\int\left(x^{2}+\frac{13 x+6}{x^{3}-x^{2}-6 x}\right) d x=\int x^{2} d x+\int \frac{13 x+6}{x(x-3)(x+2)} d x=
$$

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

$$
\begin{aligned}
& \frac{13 x+6}{x(x-3)(x+2)}=\frac{A}{x}+\frac{B}{x-3}+\frac{C}{x+2}= \\
& =\frac{A(x-3)(x+2)+B x(x+2)+C x(x-3)}{x(x-3)(x+2)}= \\
& =\frac{A\left(x^{2}-x-6\right)+B\left(x^{2}+2 x\right)+C\left(x^{2}-3 x\right)}{x(x-3)(x+2)}= \\
& =\frac{(A+B+C) x^{2}+(-A+2 B-3 C) x-6 A}{x(x-3)(x+2)}= \\
& \left\{\begin{array}{l}
A+B+C=0 \\
-A+2 B-3 C=13 \\
-6 A=6
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=1 \\
2 B-3 C=12 \\
A=-1
\end{array}\right.
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=1 \\
5 B=15 \\
A=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
C=-2 \\
B=3 \\
A=-1
\end{array}\right. \\
& \frac{13 x+6}{x(x-3)(x+2)}=-\frac{1}{x}+\frac{3}{x-3}-\frac{2}{x+2}
\end{aligned}
$$

Then we get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-27$ »

Categories: Kuznetsov's Problem Book Integrals Problem $5 \mid$ Integrals

- Last edited: 21:36, April 6, 2009.
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## Problem Kuznetsov Integrals 5-28

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