# Integral of function

Find $\displaystyle\int {1\over s^2 (s-1)^2}\,ds$.

I'm not sure how to set the integral to something like $A/s^2+B/(s-1)^2$. I don't know when do we use $Ax$ and when do we use $Ax^2$ and when do we just use $A$.

You maybe need to decompose it as $\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s-1)}+\frac{D}{(s-1)^2}$ –  Yimin Feb 15 '13 at 3:41
How about \begin{aligned}\frac{1}{s^2 (s-1)^2}&=\left(\frac{1}{s-1}-\frac{1}{s}\right)^2=\frac{1}{s^2}+\frac{1}{(s-1)^2}-\frac{2}{s(s-1)}\\&=\frac{1}{s^2}+\frac{1}{(s-1)^2}-2\left(\frac{1}{s-1}-\frac{1}{s}\right)\end{aligned} This gives $$\int\frac{1}{s^2 (s-1)^2}ds=-\frac{1}{s}-\frac{1}{s-1}-2\log\frac{|s-1|}{|s|}+C$$