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.
 A: One more way: expansion in partial fractions:
$$
\frac{1}{(a-u^2)^2}=\frac{1}{((\sqrt{a}-u)(\sqrt{a}+u))^2}= \frac{1}{4a} \bigg(\frac{1}{(\sqrt{a}-u)^2}+\frac{1}{\sqrt{a}(\sqrt{a}-u)}+\frac{1}{(\sqrt{a}+u)^2}+\frac{1}{\sqrt{a}(\sqrt{a}+u)} \bigg)
$$
so you end up with four simple integrals. Can you handle from here? 
A: Hint: Substitute $u=\sqrt{a} \sin{x}$, $du= \sqrt{a} \cos{x} dx$.
Upon doing the substitution, you should get
$$\int \frac{du}{(a-u^2)^2} = a^{-3/2} \int dx \: \sec^3{x}$$
For the latter integral, some trig identities will be needed.
$$\begin{align}\int dx \: \sec^3{x} &= \int d(\tan{x}) \sec{x}\\ &= \tan{x} \sec{x} - \int dx \: \sec{x} \tan^2{x}\\ &= \tan{x} \sec{x} + \int dx \: \sec{x} - \int dx \: \sec^3{x} \end{align}$$
So now
$$\begin{align}2\int dx \: \sec^3{x} &= \tan{x} \sec{x} + \int dx \: \sec{x}\end{align}$$
Can you do the rest yourself?
A: HINT: First substitute $u=\sqrt a \sin x$ and then $\tan x =t$
. These will convert it into a standard integral.
A: 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)} $
