## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{4}+2 x^{3}-3 x^{2}+2 x-9}{x(x-1)(x+3)} d x
$$
## Solution

$$
\int \frac{2 x^{4}+2 x^{3}-3 x^{2}+2 x-9}{x(x-1)(x+3)} d x=\int \frac{2 x^{4}+2 x^{3}-3 x^{2}+2 x-9}{x^{3}+2 x^{2}-3 x} d x=
$$

Under the integral, we have an improper fraction. Let's separate the integer part:

$$
\begin{array}{ll}
2 x^{4}+2 x^{3}-3 x^{2}+2 x-9 & \frac{\mid x^{3}+2 x^{2}-3 x}{2 x-2} \\
\frac{2 x^{4}+4 x^{3}-6 x^{2}}{-2 x^{3}+3 x^{2}+2 x-9} & \\
\quad \frac{-2 x^{3}-4 x^{2}+6 x}{7 x^{2}-4 x-9} &
\end{array}
$$

We get:

$$
=\int\left(2 x-2+\frac{7 x^{2}-4 x-9}{x^{3}-3 x^{2}+2 x}\right) d x=\int(2 x-2) d x+\int \frac{7 x^{2}-4 x-9}{x(x-1)(x+3)} d x=
$$

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

$$
\begin{aligned}
& \frac{7 x^{2}-4 x-9}{x(x-1)(x+3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+3}= \\
& =\frac{A(x+3)(x-1)+B(x+3) x+C(x-1) x}{x(x-1)(x+3)}= \\
& =\frac{A\left(x^{2}+2 x-3\right)+B\left(x^{2}+3 x\right)+C\left(x^{2}-x\right)}{x(x-1)(x+3)}= \\
& =\frac{(A+B+C) x^{2}+(2 A+3 B-C) x-3 A}{x(x-1)(x+3)}= \\
& \left\{\begin{array}{l}
A+B+C=7 \\
2 A+3 B-C=-4 \\
-3 A=-9
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=4 \\
3 B-C=-10 \\
A=3
\end{array}\right.
$$

Add the first row to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=4 \\
4 B=-6 \\
A=3
\end{array}\right. \\
& \left\{\begin{array}{l}
C=\frac{11}{2} \\
B=-\frac{3}{2} \\
A=3
\end{array}\right. \\
& \frac{3 x^{2}-4 x+9}{(x+3)(x-1) x}=\frac{3}{x}-\frac{3}{2(x-1)}+\frac{11}{2(x+3)}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int(2 x-2) d x+\int\left(\frac{3}{x}-\frac{3}{2(x-1)}+\frac{11}{2(x+3)}\right) d x= \\
& =x^{2}-2 x+3 \cdot \ln |x|-\frac{3}{2} \cdot \ln |x-1|+\frac{11}{2} \cdot \ln |x+3|+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-29$ ”

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

- Last edited: 12:00, April 7, 2009.
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## Problem Kuznetsov Integrals 5-30

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