Find the value of $\int \frac {du}{(a-u^2)^2}$ I am stuck on the following integration problem:   

$\int \frac {du}{(a-u^2)^2}; a$ being a constant.   

Can someone point me in the right direction? Thanks in advance for your time.

Use $u = \sqrt{a} \sin{x}$; we get:
$$
\begin{align}
\int{\dfrac{du}{ (a - u^2)^2}} &= \int{\dfrac{\cos{x} dx}{ (a - a\sin^2{x})^2}} \\
&= \dfrac{1}{a^2}\int{\dfrac{\cos{x} dx}{ \cos^4{x}}} \\
&= \dfrac{1}{a^2}\int{\sec^3{x} dx} \\
&= \dfrac{1}{a^2} \cdot \left( \dfrac{1}{2} \sec{x} \tan{x} + \dfrac{1}{2} \ln{|\sec{x} + \tan{x} |}\right) + C
\end{align}
$$
Where, I used this method to integrate $\sec^3{x}$.
Substitute back the value of $x = \sin^{-1}{ \left(\dfrac{u}{\sqrt{a}} \right)} $
