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

Seja $x = 2\tan \theta$, $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. $dx = 2\sec^2 \theta\ d\theta$

$I\ =\ \displaystyle\int \sec^3 \theta\ d\theta\ =\ (\sec \theta)(\tan \theta) - \displaystyle\int (\sec \theta)(\tan^2 \theta)\ d\theta\ =$

$=\ (\sec \theta)(\tan \theta) - \displaystyle\int \sec^3 \theta\ d\theta + \log |\sec \theta + \tan \theta|\ \Rightarrow$

$\Rightarrow\ I = \dfrac{(\sec \theta)(\tan \theta) + \log |\sec \theta + \tan \theta|}{2} + c\ =\ \fbox{$\dfrac{x\sqrt{4 + x^2}}{8} + \dfrac{\log \left|\sqrt{4 + x^2} + x\right|}{2} + c$}$