Evaluate the following integral:
$\int \frac{18}{(x+2)\left(x^{2}+4\right)} d x$
$I=\int \frac{18}{(x+2)\left(x^{2}+4\right)}$
$\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+4}$
$18=A\left(x^{2}+4\right)+(B x+C)(x+2)$
Equating constants
$18=4 \mathrm{~A}+2 \mathrm{C}$
Equating coefficients of $x$
$0=2 B+C$
Equating coefficients of $x^{2}$
$0=A+B$
Solving, we get
$A=\frac{9}{4}, \quad B=-\frac{9}{4}, \quad C=\frac{9}{2}$
Thus,
$I=\frac{9}{4} \int \frac{d x}{x+2}+\left(-\frac{9}{4}\right) \int \frac{x d x}{x^{2}+4}+\frac{9}{2} \int \frac{d x}{x^{2}+4}$
$I=\frac{9}{4} \log |x+2|-\frac{9}{8} \log \left|x^{2}+4\right|+\frac{9}{4} \tan ^{-1}\left(\frac{x}{2}\right)+C$
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