## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}} d x
$$
## Solution

$$
\int \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3}}{(x+1)^{3}}= \\
& =\frac{A(x+1)^{3}+B_{1}(x-2)(x+1)^{2}+B_{2}(x-2)(x+1)+B_{3}(x-2)}{(x-2)(x+1)^{3}}= \\
& =\frac{A\left(x^{3}+3 x^{2}+3 x+1\right)+B_{1}\left(x^{3}-3 x-2\right)+B_{2}\left(x^{2}-x-2\right)+B_{3}(x-2)}{(x-2)(x+1)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(3 A+B_{2}\right) x^{2}+\left(3 A-3 B_{1}-B_{2}+B_{3}\right) x+\left(A-2 B_{1}-2 B_{2}-2 B_{3}\right)}{(x-2)(x+1)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=2 \\
3 A+B_{2}=6 \\
3 A-3 B_{1}-B_{2}+B_{3}=5 \\
A-2 B_{1}-2 B_{2}-2 B_{3}=4
\end{array}\right.
\end{aligned}
$$

Add the first equation multiplied by -3 to the second equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=2 \\
-3 B_{1}+B_{2}=0 \\
3 A-3 B_{1}-B_{2}+B_{3}=5 \\
A-2 B_{1}-2 B_{2}-2 B_{3}=4
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=2 \\
B_{2}=3 B_{1} \\
3 A-6 B_{1}+B_{3}=5 \\
A-8 B_{1}-2 B_{3}=4
\end{array}\right.
\end{aligned}
$$

Add the third equation multiplied by 2 to the fourth equation:

$$
\left\{\begin{array}{l}
A+B_{1}=2 \\
B_{2}=3 B_{1} \\
3 A-6 B_{1}+B_{3}=5 \\
7 A-20 B_{1}=14
\end{array}\right.
$$

Add the first equation multiplied by 20 to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=2 \\
B_{2}=3 B_{1} \\
3 A-6 B_{1}+B_{3}=5 \\
27 A=54
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=3 B_{1} \\
-6 B_{1}+B_{3}=-1 \\
A=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=-1 \\
A=2
\end{array}\right. \\
& \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}}=\frac{2}{x-2}-\frac{1}{(x+1)^{3}}
\end{aligned}
$$

Then

$$
\int \frac{2 x^{3}+6 x^{2}+5 x+4}{(x-2)(x+1)^{3}} d x=\int\left(\frac{2}{x-2}-\frac{1}{(x+1)^{3}}\right) d x=2 \cdot \ln |x-2|+\frac{1}{2(x+1)^{2}}+C
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_6-20$ » Categories: Kuznetsov's Integral Problems 6 | Integrals

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- Last edited: 09:59, April 9, 2009.
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## Problem Kuznetsov Integrals 6-21

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