## Problem Statement

Find the indefinite integral:

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

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

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

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

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

$$
\left\{\begin{array}{l}
A+C=0 \\
B+C+D=0 \\
D=1 \\
2 B+2 D=4
\end{array}\right.
$$

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

Then:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-28$ »

Categories: Kuznetsov's Problem Book Integrals Problem 7 | Integrals

- Last edited: 09:14, 22 April 2009.
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## Problem Kuznetsov Integrals $7-29$

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