How can I solve this integral with complex number $$\int_{|z-1| = 1}\frac{1}{(1-z^2)}dz$$
I tried to do this by residue calculus
 $$I=2\pi*iRes(f(z),1)$$
but I coudn't get the answer..
I would be grateful if you could give a clue.
Additional question) 
$$\int_{|z| = 3}\frac{z}{(1-z^2)}dz$$
Is $$I=2\pi*i{Res(f,0)}+Res(f,1)$$ and the answer is 0 right?

HINT:
Note that 
$$\frac{1}{1-z^2}=\frac{1/2}{1-z}+ \frac{1/2}{1+z}$$
The poles are at $z=\pm 1$ and the residues are $\pm 1/2$.  Now, which poles, if any are contained in $|z-1|<1$?  Which are contained in $|z-1| <3$? 
