Integrate the following integral using partial fractions I want to integrate 
$$\frac{1}{(1-u^2)^2}. $$
I have used the difference of two squares to get
$$\frac{1}{2(1-u^2)} + \frac{1}{2(1+u^2)} $$
and then integrated it to get
$$\frac{1}{2ln|1-u^2|} + \frac{1}{2ln|1+u^2|}.$$

Just wondering if this is correct? thanks

 A: If you want to use partial fraction decomposition, then note that: $$\dfrac 1{(1-u^2)^2} = \dfrac 1{[(u-1)(u+1)]^2} = \dfrac 1{(u-1)^2(u+1)^2} $$
$$= \dfrac{A}{(u-1)} + \dfrac{B}{(u -1)^2} + \dfrac{C}{(u+1)} + \dfrac D{(u+1)^2}$$
Now try solving for the needed constant terms: $A,\, B, \,C,\, D$.
Spoiler I (to check your solutions to $A, B, C, D$) 

 $$A = -\dfrac 14, \;B = C = D = \dfrac 14.\quad$$ 

Spoiler II

 $$\int \dfrac {du}{(1-u^2)^2} = \dfrac 14 \left(\int \dfrac{-du}{(u-1)} + \int \dfrac{du}{(u -1)^2} + \int \dfrac{du}{(u+1)} + \dfrac {du}{(u+1)^2}\right)\\ \\ = \dfrac 14 \left(-\ln|u - 1| - \dfrac{1}{u - 1} + \ln|u+1| - \dfrac{1}{u+1}\right) + C \\ \\ = \dfrac 14\left( \ln \left|\dfrac{u+1}{u-1}\right| - \dfrac{2u}{u^2 - 1}\right) + C$$

A: Hint: It is easier to use the substitution $u=\sin(x)$.
A: Use substitution $u=\sin x \implies \mathrm{d}u=\cos x \ \mathrm{d}x$. The integral then  becomes:
$$\int \frac{1}{(1-u^2)^2}\ \mathrm{d}u=\int\frac{\cos x \ \mathrm{d}x}{(\cos^2x)^2}=\int\frac{ \mathrm{d}x}{(\cos x)^3}=\int (\sec x)^3\mathrm{d}x$$
The integral of the secant cubed is detailed here
