## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{3 x^{3}-2}{x^{3}-x} d x
$$
## Solution

$$
\int \frac{3 x^{3}-2}{x^{3}-x} d x=
$$

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

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

We get:

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

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

$$
\begin{aligned}
& \frac{3 x-2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}= \\
& =\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)}= \\
& =\frac{A\left(x^{2}-1\right)+B\left(x^{2}+x\right)+C\left(x^{2}-x\right)}{x(x-1)(x+1)}= \\
& =\frac{(A+B+C) x^{2}+(B-C) x-A}{x(x-1)(x+1)}= \\
& \left\{\begin{array}{l}
A+B+C=0 \\
B-C=3 \\
-A=-2
\end{array}\right.
\end{aligned}
$$

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

Add the first row to the second:

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

Then we get:

$$
\begin{aligned}
& =\int 3 d x+\int\left(\frac{2}{x}+\frac{1}{2(x-1)}-\frac{5}{2(x+1)}\right) d x= \\
& =3 x+2 \cdot \ln |x|+\frac{1}{2} \cdot \ln |x-1|-\frac{5}{2} \cdot \ln |x+1|+C
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_5-14" Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

Ukrainian Banner Network

- Last modified: 08:55, 6 April 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 5-14

## Material from Plusi