## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}-1}{x^{2}+x-6} d x
$$
## Solution

$$
\int \frac{2 x^{3}-1}{x^{2}+x-6} d x=
$$

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

$$
\begin{array}{ll}
2 x^{3}-1 & \frac{x^{2}+x-6}{2 x-2} \\
\frac{2 x^{3}+2 x^{2}-12 x}{-2 x^{2}+12 x-1}-1 & \\
\frac{-2 x^{2}-2 x+12}{14 x-13} &
\end{array}
$$

We get:

$$
=\int\left(2 x-2+\frac{14 x-13}{x^{2}+x-6}\right) d x=\int(2 x-2) d x+\int \frac{14 x-13}{(x+3)(x-2)} d x=
$$

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

$$
\begin{aligned}
& \frac{14 x-13}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}=\frac{A(x-2)+B(x+3)}{(x+3)(x-2)}=\frac{(A+B) x+(-2 A+3 B)}{(x+3)(x-2)} \\
& \left\{\begin{array}{l}
A+B=14 \\
-2 A+3 B=-13
\end{array}\right. \\
& \left\{\begin{array}{l}
B=3 \\
A=11
\end{array}\right. \\
& \frac{14 x-13}{(x+3)(x-2)}=\frac{11}{x+3}+\frac{3}{x-2}
\end{aligned}
$$

Then we get:

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

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

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

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## Problem Kuznetsov Integrals 5-6

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