## Problem Statement

Find the indefinite integral:

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

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

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

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

Subtract the second equation multiplied by 2 from the third equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=3 \\
A_{1}+A_{2}+2 B+C=6 \\
-2 A_{2}-3 B=-7 \\
2 A_{1}+2 A_{2}+C=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=3-A_{1} \\
A_{1}+A_{2}+6-2 A_{1}+C=6 \\
-2 A_{2}-9+3 A_{1}=-7 \\
2 A_{1}+2 A_{2}+C=-1
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
B=3-A_{1} \\
A_{2}-A_{1}+C=0 \\
3 A_{1}-2 A_{2}=2 \\
C=-1-2 A_{1}-2 A_{2}
\end{array}\right. \\
& \left\{\begin{array}{l}
B=3-A_{1} \\
-3 A_{1}-A_{2}=1 \\
3 A_{1}-2 A_{2}=2 \\
C=-1-2 A_{1}-2 A_{2}
\end{array}\right.
\end{aligned}
$$

Add the second equation to the third equation:

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

Then:

$$
\begin{aligned}
& \int \frac{3 x^{3}+6 x^{2}+5 x-1}{(x+1)^{2}\left(x^{2}+2\right)} d x=\int\left(-\frac{1}{(x+1)^{2}}+\frac{3 x+1}{x^{2}+2}\right) d x= \\
& =-\int \frac{1}{(x+1)^{2}} d x+\frac{3}{2} \cdot \int \frac{2 x+\frac{2}{3}}{x^{2}+2} d x= \\
& =\frac{1}{x+1}+\frac{3}{2} \cdot \int \frac{2 x}{x^{2}+2} d x+\frac{3}{2} \cdot \frac{2}{3} \cdot \int \frac{1}{x^{2}+2} d x= \\
& =\frac{1}{x+1}+\frac{3}{2} \cdot \int \frac{d\left(x^{2}+2\right)}{x^{2}+2}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{x}{\sqrt{2}}= \\
& =\frac{1}{x+1}+\frac{3}{2} \cdot \ln \left(x^{2}+2\right)+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{x}{\sqrt{2}}+C
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-7$ >

Categories: Kuznetsov's Integral Problems 7 | Integrals

- Last edited: 09:45, 2 May 2009.
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## Problem Kuznetsov Integrals $7-8$

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