Need help calculating integral $\int\frac{dx}{(x^2+4)^2}$

Question

I need some help calculating the integral

$\int\frac{dx}{(x^2+4)^2}$

I tried integration by parts, but I could not arrive to an answer.

Answer 1

Try letting $x=2\tan t$. Then $x^2+4=4(1+\tan^2 t)=4\sec^2 t$, $dx=2\sec^2 tdt$.

Answer 2

As an alternative, note that

$$\int \frac{dx}{(x^2+a^2)^2} = -\frac{1}{2 a} \frac{\partial}{\partial a} \int \frac{dx}{x^2+a^2} = -\frac{1}{2 a} \frac{\partial}{\partial a} \frac{\arctan{(x/a)}}{a}$$

Carry out the differentiation and substitute $a=2$; I get

$$\frac18 \frac{x}{x^2+4} + \frac{1}{16} \arctan{\frac{x}{2}}+C$$