## Problem Statement

Calculate the indefinite integral:

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

$$
\int \frac{2 x^{3}+5}{x^{2}-x-2} d x=
$$

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

$$
\begin{array}{ll}
\frac{2 x^{3}+5}{2 x^{3}-2 x^{2}-4 x} & \frac{\mid x^{2}-x-2}{2 x+2} \\
2 x^{2}+4 x+5 & \\
\frac{2 x^{2}-2 x-4}{6 x+9} &
\end{array}
$$

We get:

$$
=\int\left(2 x+2+\frac{6 x+9}{x^{2}-x-2}\right) d x=\int(2 x+2) d x+\int \frac{6 x+9}{(x-2)(x+1)} d x=
$$

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

$$
\begin{aligned}
& \frac{6 x+9}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}=\frac{(A+B) x+(A-2 B)}{(x-2)(x+1)} \\
& \left\{\begin{array}{l}
A+B=6 \\
A-2 B=9
\end{array}\right. \\
& \left\{\begin{array}{l}
A=7 \\
B=-1
\end{array}\right. \\
& \frac{6 x+9}{(x-2)(x+1)}=\frac{7}{x-2}-\frac{1}{x+1}
\end{aligned}
$$

Then we get:

$$
=\int(2 x+2) d x+\int\left(\frac{7}{x-2}-\frac{1}{x+1}\right) d x=x^{2}+2 x+7 \ln |x-2|-\ln |x+1|+C
$$

% 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% 20% D0% B7% D0% B0% D0% B4% D0% B0% D1% 87% D0% B0% 205-4

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

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## Problem Kuznetsov Integrals 5-5

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