## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{3 x^{4}+3 x^{3}-5 x^{2}+2}{x(x-1)(x+2)} d x
$$
## Solution

$$
\int \frac{3 x^{4}+3 x^{3}-5 x^{2}+2}{x(x-1)(x+2)} d x=\int \frac{3 x^{4}+3 x^{3}-5 x^{2}+2}{x^{3}+x^{2}-2 x} d x=
$$

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

$$
\begin{array}{ll}
3 x^{4}+3 x^{3}-5 x^{2}+2 & \mid x^{3}+x^{2}-2 x \\
\frac{3 x^{4}+3 x^{3}-6 x^{2}}{x^{2}+2} & 3 x
\end{array}
$$

We get:

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

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

$$
\begin{aligned}
& \frac{x^{2}+2}{x(x-1)(x+2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}= \\
& =\frac{A(x-1)(x+2)+B x(x+2)+C x(x-1)}{x(x-1)(x+2)}= \\
& =\frac{A\left(x^{2}+x-2\right)+B\left(x^{2}+2 x\right)+C\left(x^{2}-x\right)}{x(x-1)(x+2)}= \\
& =\frac{(A+B+C) x^{2}+(A+2 B-C) x-2 A}{x(x-1)(x+2)}= \\
& \left\{\begin{array}{l}
A+B+C=1 \\
A+2 B-C=0 \\
-2 A=2
\end{array}\right.
\end{aligned}
$$

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

Add the first row to the second:

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

Then we get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-25$ "

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

Ukrainian Banner Network

- Last modified: 21:04, 6 April 2009.
- Content is available under CC-BY-SA 3.0.