solve the integral bu substitution I tried substituting $x=sin\ u$ but I didn't get nowhere, can someone just give me a hint how to solve this integral?

$$\int\frac{dx}{(x^2-1)^2}$$

 A: Use partial fractions. We have
$$\dfrac1{(x^2-1)^2} = \dfrac14\left(\dfrac1{x+1} + \dfrac1{(x+1)^2} - \dfrac1{x-1} + \dfrac1{(x-1)^2}\right)$$
Each of the term on the right hand side can be integrated trivially.
A: The integration can be accomplished using the substitution that you favor, but the partial fraction decomposition is a much simpler method.  
Let $x=sin\ y$.
$\ \ \ \ dx = cos\ y\ dy$
 $$\displaystyle\int\frac{dx}{(x^2-1)^2}=\displaystyle\int\frac{dx}{(1-x^2)^2}=\displaystyle\int\frac{cos\ y\ dy}{(1-sin^2\ y)^2}=\displaystyle\int\frac{cos\ y\ dy}{(cos^2\ y)^2}=\displaystyle\int sec^3\ y\ dy$$
From here, it can be solved using integration by parts:  
Put $\ \ \ \ u=sec\ y\ \ \ \ \ \ \ \ \ \ \ \ \  $ and $\ \ \ \ \ \ \ dv=sec^2\ y$
$\ \ \ \ \  \ du=sec\ y\ tan\ y$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v=tan\ y$  
Then $\ \displaystyle\int sec^3\ y\ dy\ =\ \displaystyle\int (u\ dv)\ dy$
Afterward, can recover the original integral by substituting $y=arcsin\ x$.  
Seems like a lot of work to do it this way.
