## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}} d x=
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}}=\frac{A}{x-1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x+2)^{3}}= \\
& =\frac{A(x+2)^{3}+B_{1}(x-1)(x+2)^{2}+B_{2}(x-1)(x+2)+B_{3}(x-1)}{(x-1)(x+2)^{3}}= \\
& =\frac{A\left(x^{3}+6 x^{2}+12 x+8\right)+B_{1}\left(x^{3}+3 x^{2}-4\right)+B_{2}\left(x^{2}+x-2\right)+B_{3}(x-1)}{(x-1)(x+2)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(6 A+3 B_{1}+B_{2}\right) x^{2}+\left(12 A+B_{2}+B_{3}\right) x+\left(8 A-4 B_{1}-2 B_{2}-B_{3}\right)}{(x-1)(x+2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+3 B_{1}+B_{2}=6 \\
12 A+B_{2}+B_{3}=10 \\
8 A-4 B_{1}-2 B_{2}-B_{3}=10
\end{array}\right.
\end{aligned}
$$

Add the third equation to the fourth:

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+3 B_{1}+B_{2}=6 \\
12 A+B_{2}+B_{3}=10 \\
20 A-4 B_{1}-B_{2}=20
\end{array}\right.
$$

Add the second equation to the fourth:

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+3 B_{1}+B_{2}=6 \\
12 A+B_{2}+B_{3}=10 \\
26 A-B_{1}=26
\end{array}\right.
$$

Add the first equation to the fourth:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+3 B_{1}+B_{2}=6 \\
12 A+B_{2}+B_{3}=10 \\
27 A=27
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
3 B_{1}+B_{2}=0 \\
B_{2}+B_{3}=-2 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=-2 \\
A=1
\end{array}\right. \\
& \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}}=\frac{1}{x-1}-\frac{2}{(x+2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}} d x=\int\left(\frac{1}{x-1}-\frac{2}{(x+2)^{3}}\right) d x=\ln |x-1|+\frac{1}{(x+2)^{2}}+C
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_6-8"

Categories: Kuznetsov's Integral Problems 6 | Integrals

- Last modified: 08:53, 8 April 2009.
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## Problem Kuznetsov Integrals 6-9

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