Calcular $I\ =\ \displaystyle\int \dfrac{x}{x^2 + 4}\ dx$.

Seja $x = 2\tan u$, $dx = 2\sec^2 u\ du$.

$I\ =\ \displaystyle\int \tan u\ du\ =\ \displaystyle\int \dfrac{\sin u}{\cos u} du$

Seja $v = \cos u$, $dv = -\sin u\ du$.

$I\ =\ -\displaystyle\int \dfrac{dv}{v} = -(\log |v|) + c = -(\log |\cos u|) + c = \fbox{$-\left(\log \cos \arctan \dfrac{x}{2}\right) + c$}$