## Problem Statement

Find the indefinite integral:

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

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

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

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

Subtract the first equation from the third:

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

Then:

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 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}+7-22 »

Categories: Kuznetsov's Integral Problems 7 | Integrals

- Last edited: 16:24, 21 April 2009.
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Problem Kuznetsov Integrals 7-22 — PlusPi

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## Problem Kuznetsov Integrals $7-23$

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