Evaluate the following integral:

$I=\int \frac{18}{(x+2)\left(x^{2}+4\right)}$

$\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+4}$

$18=A\left(x^{2}+4\right)+(B x+C)(x+2)$

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,

$I=\frac{9}{4} \int \frac{d x}{x+2}+\left(-\frac{9}{4}\right) \int \frac{x d x}{x^{2}+4}+\frac{9}{2} \int \frac{d x}{x^{2}+4}$

$I=\frac{9}{4} \log |x+2|-\frac{9}{8} \log \left|x^{2}+4\right|+\frac{9}{4} \tan ^{-1}\left(\frac{x}{2}\right)+C$