## Problem Statement

Find the indefinite integral:

$$
\int \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)} d x
$$
## Solution

$$
\int \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)}=\frac{A_{1}}{x+2}+\frac{A_{2}}{(x+2)^{2}}+\frac{B x+C}{x^{2}+4}= \\
& =\frac{A_{1}(x+2)\left(x^{2}+4\right)+A_{2}\left(x^{2}+4\right)+(B x+C)(x+2)^{2}}{(x+2)^{2}\left(x^{2}+4\right)}= \\
& =\frac{A_{1}\left(x^{3}+2 x^{2}+4 x+8\right)+A_{2}\left(x^{2}+4\right)+B\left(x^{3}+4 x^{2}+4 x\right)+C\left(x^{2}+4 x+4\right)}{(x+2)^{2}\left(x^{2}+4\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(2 A_{1}+A_{2}+4 B+C\right) x^{2}+\left(4 A_{1}+4 B+4 C\right) x+\left(8 A_{1}+4 A_{2}+4 C\right)}{(x+3)^{2}\left(x^{2}+3\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=1 \\
2 A_{1}+A_{2}+4 B+C=6 \\
4 A_{1}+4 B+4 C=8 \\
8 A_{1}+4 A_{2}+4 C=8
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=1 \\
2 A_{1}+A_{2}+4 B+C=6 \\
A_{1}+B+C=2 \\
2 A_{1}+A_{2}+C=2
\end{array}\right.
\end{aligned}
$$

Subtract the second equation from the fourth:

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}=0 \\
2 A_{1}+A_{2}+C=2 \\
A_{1}+C=1 \\
B=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}+C=2 \\
C=1 \\
B=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}=1 \\
C=1 \\
B=1
\end{array}\right. \\
& \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)}=\frac{1}{(x+2)^{2}}+\frac{x+1}{x^{2}+4}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)} d x=\int\left(\frac{1}{(x+2)^{2}}+\frac{x+1}{x^{2}+4}\right) d x= \\
& =\int \frac{1}{(x+2)^{2}} d x+\frac{1}{2} \cdot \int \frac{2 x+2}{x^{2}+4} d x= \\
& =-\frac{1}{x+2}+\frac{1}{2} \cdot \int \frac{2 x}{x^{2}+4} d x+\frac{1}{2} \cdot \int \frac{2}{x^{2}+4} d x= \\
& =-\frac{1}{x+2}+\frac{1}{2} \cdot \int \frac{d\left(x^{2}+4\right)}{x^{2}+4}+\int \frac{1}{x^{2}+4} d x= \\
& =-\frac{1}{x+2}+\frac{1}{2} \cdot \ln \left(x^{2}+4\right)+\frac{1}{2} \cdot \operatorname{arctg} \frac{x}{2}+C
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-9$ "

Categories: Kuznetsov's Integral Problems 7 | Integrals

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## Problem Kuznetsov Integrals 7-10

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