## Problem Statement

Find the indefinite integral:

$$
\int \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)} d x
$$
## Solution

$$
\int \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)}=\frac{A_{1}}{x-1}+\frac{A_{2}}{(x-1)^{2}}+\frac{B x+C}{x^{2}+4 x+5}= \\
& =\frac{A_{1}(x-1)\left(x^{2}+4 x+5\right)+A_{2}\left(x^{2}+4 x+5\right)+(B x+C)(x-1)^{2}}{(x-1)^{2}\left(x^{2}+4 x+5\right)}= \\
& =\frac{A_{1}\left(x^{3}+3 x^{2}+x-5\right)+A_{2}\left(x^{2}+4 x+5\right)+B\left(x^{3}-2 x^{2}+x\right)+C\left(x^{2}-2 x+1\right)}{(x-1)^{2}\left(x^{2}+4 x+5\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(3 A_{1}+A_{2}-2 B+C\right) x^{2}+\left(A_{1}+4 A_{2}+B-2 C\right) x+\left(-5 A_{1}+5 A_{2}+C\right)}{(x-1)^{2}\left(x^{2}+4 x+5\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=2 \\
3 A_{1}+A_{2}-2 B+C=-4 \\
A_{1}+4 A_{2}+B-2 C=-16 \\
-5 A_{1}+5 A_{2}+C=-12
\end{array}\right.
\end{aligned}
$$

Add the first equation multiplied by 2 to the second equation:

$$
\left\{\begin{array}{l}
A_{1}+B=2 \\
5 A_{1}+A_{2}+C=0 \\
A_{1}+4 A_{2}+B-2 C=-16 \\
-5 A_{1}+5 A_{2}+C=-12
\end{array}\right.
$$

Add the second equation to the fourth equation:

$$
\left\{\begin{array}{l}
A_{1}+B=2 \\
5 A_{1}+A_{2}+C=0 \\
A_{1}+4 A_{2}+B-2 C=-16 \\
6 A_{2}+2 C=-12
\end{array}\right.
$$

Add the fourth equation to the third equation:

$$
\left\{\begin{array}{l}
A_{1}+B=2 \\
5 A_{1}+A_{2}+C=0 \\
A_{1}+10 A_{2}+B=-28 \\
6 A_{2}+2 C=-12
\end{array}\right.
$$

Subtract the first equation from the third equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=2 \\
5 A_{1}+C=3 \\
10 A_{2}=-30 \\
6 A_{2}+2 C=-12
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=2 \\
5 A_{1}=0 \\
A_{2}=-3 \\
2 C=6
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=2 \\
5 A_{1}=0 \\
A_{2}=-3 \\
C=3 \\
B=2 \\
A_{1}=0 \\
A_{2}=-3 \\
C=3
\end{array}\right. \\
& \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)}=-\frac{3}{(x-1)^{2}}+\frac{2 x+3}{x^{2}+4 x+5}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)} d x=\int\left(-\frac{3}{(x-1)^{2}}+\frac{2 x+3}{x^{2}+4 x+5}\right) d x= \\
& =-\int \frac{3}{(x-1)^{2}} d x+\int \frac{2 x+4-1}{x^{2}+4 x+5} d x= \\
& =\frac{3}{x-1}+\int \frac{2 x+4}{x^{2}+4 x+5} d x-\int \frac{1}{x^{2}+4 x+5} d x= \\
& =\frac{3}{x-1}+\int \frac{d\left(x^{2}+4 x+5\right)}{x^{2}+4 x+5}-\int \frac{1}{(x+2)^{2}+1} d x= \\
& =\frac{3}{x-1}+\ln \left(x^{2}+4 x+5\right)-\operatorname{arctg}(x+2)+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-11 »

Categories: Kuznetsov's Integral Problems 7 | Integrals

- Last edited: 19:37, April 20, 2009.
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Kuznetsov's Integral Problems 7-11 — PlusPi
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## Kuznetsov's Integral Problems $7-12$

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