## Problem Statement

Find the indefinite integral:

$$
\int \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x
$$
## Solution

$$
\int \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)^{2}}+\frac{B x+C}{x^{2}+2 x+2}= \\
& =\frac{A_{1}(x+1)\left(x^{2}+2 x+2\right)+A_{2}\left(x^{2}+2 x+2\right)+(B x+C)(x+1)^{2}}{(x+1)^{2}\left(x^{2}+2 x+2\right)}= \\
& =\frac{A_{1}\left(x^{3}+3 x^{2}+4 x+2\right)+A_{2}\left(x^{2}+2 x+2\right)+B\left(x^{3}+2 x^{2}+x\right)+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}\left(x^{2}+2 x+2\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(3 A_{1}+A_{2}+2 B+C\right) x^{2}+\left(4 A_{1}+2 A_{2}+B+2 C\right) x+\left(2 A_{1}+2 A_{2}+C\right)}{(x+1)^{2}\left(x^{2}+2 x+2\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=1 \\
3 A_{1}+A_{2}+2 B+C=6 \\
4 A_{1}+2 A_{2}+B+2 C=9 \\
2 A_{1}+2 A_{2}+C=6
\end{array}\right.
\end{aligned}
$$

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

$$
\left\{\begin{array}{l}
A_{1}+B=1 \\
A_{1}+A_{2}+C=4 \\
4 A_{1}+2 A_{2}+B+2 C=9 \\
2 A_{1}+2 A_{2}+C=6
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
A_{1}+B=1 \\
A_{1}+A_{2}+C=4 \\
4 A_{1}+2 A_{2}+B+2 C=9 \\
-C=-2
\end{array}\right.
$$

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

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

$$
\left\{\begin{array}{l}
A_{1}+B=1 \\
A_{1}+A_{2}=2 \\
2 A_{1}+B=1 \\
C=2
\end{array}\right.
$$

Subtract the first equation from the third equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=1 \\
A_{1}+A_{2}=2 \\
A_{1}=0 \\
C=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1 \\
A_{2}=2 \\
A_{1}=0 \\
C=2
\end{array}\right. \\
& \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)}=\frac{2}{(x+1)^{2}}+\frac{x+2}{x^{2}+2 x+2}
\end{aligned}
$$

Then

$$
\begin{aligned}
& \int \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x=\int\left(\frac{2}{(x+1)^{2}}+\frac{x+2}{x^{2}+2 x+2}\right) d x= \\
& =\int \frac{2}{(x+1)^{2}} d x+\frac{1}{2} \cdot \int \frac{2 x+2+2}{x^{2}+2 x+2} d x= \\
& =-\frac{2}{x+1}+\frac{1}{2} \cdot \int \frac{2 x+2}{x^{2}+2 x+2} d x+\frac{1}{2} \cdot \int \frac{2}{x^{2}+2 x+2} d x= \\
& =-\frac{2}{x+1}+\frac{1}{2} \cdot \int \frac{d\left(x^{2}+2 x+2\right)}{x^{2}+2 x+2}+\int \frac{1}{(x+1)^{2}+1} d x= \\
& =-\frac{2}{x+1}+\frac{1}{2} \cdot \ln \left(x^{2}+2 x+2\right)+\operatorname{arctg}(x+1)+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_7-5$ "

Categories: Kuznetsov's Integral Problems 7 | Integrals

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## Problem Kuznetsov Integrals 7-6

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