Answer
$\frac{3}{2}\ln(x^2+4)+\tan^{-1}(x/2)+c.$
Work Step by Step
We have
$$
\int \frac{(3 x+2) d x}{x^{2}+4}=\int \frac{3 x}{x^{2}+4}d x+\int \frac{2 }{x^{2}+4}d x\\
=\frac{3}{2}\int \frac{2 x}{x^{2}+4}d x+\int \frac{1 }{(x/2)^{2}+1}d( x/2)\\
=\frac{3}{2}\ln(x^2+4)+\tan^{-1}(x/2)+c.
$$
