How to integrate fraction Hi can anybody give me some hint how to integrate this type of fractions
$$\int \frac{1}{(x^2 +a)^2} dx $$ where $ a \in \mathbb N $
Thanks
 A: Take the integral
$$\int \frac{dx}{x^2+a^2}$$
and we do an integration by parts:
$$\int 1\cdot\frac{dx}{x^2+a^2}=\frac{x}{x^2+a^2}+2\int\frac{x^2dx}{(x^2+a^2)^2}$$
for the last integral we add and subtract $1$ in the numerator and we find
$$\int\frac{dx}{(x^2+a^2)^2}=\frac 1 2\left(\int \frac{dx}{x^2+a^2}+\frac{x}{x^2+a^2}\right)$$
Notice that $$\int \frac{dx}{x^2+a^2}=\frac 1 a \arctan \left(\frac x a\right)+C$$
A: take x=√a tanø
then $$\int \frac{1}{(x^2 + a)^2} \, dx = \int \frac{1}{\sqrt{a³} \sec^2(\phi)} \, d\phi = \int \frac{\cos^2(\phi)}{\sqrt{a³}} \, d\phi.$$
now solve it
A: $\newcommand{\+}{^{\dagger}}%
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$$
\int{1 \over x^{2} +a}\,\dd x
= {1 \over \root{a}}\int{1 \over \pars{x/\root{a}}^{2} +1}\,{\dd x \over \root{a}}
=a^{-1/2}\arctan\pars{a^{-1/2}x}
$$

Derive both members respect of $a$:
$$
-\int{1 \over \pars{x^{2} +a}^2}\,\dd x=
-\,\half\,a^{-3/2}\arctan\pars{a^{-1/2}x} +
a^{-1/2}\,{-a^{-3/2}x/2 \over \pars{a^{-1/2}x}^{2} + 1}
$$

$$\color{#00f}{\large%
\int{1 \over \pars{x^{2} +a}^2}\,\dd x=
{1 \over 2a\root{a}}\,\arctan\pars{x \over \root{a}} +
{x \over 2a\pars{x^{2} + a}}} + \mbox{a constant}
$$
