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?

(1) We have $\int_{|z-1| = 1}\frac{1}{(1-z^2)}dz=2\pi iRes(f(z);1)$
$Res(f(z);1)=\displaystyle \lim_{z\to1}\frac{z-1}{(1-z^2)}=\frac {-1}2$
$\int_{|z-1| = 1}\frac{1}{(1-z^2)}dz=2\pi i\times\frac{-1}{2}=-\pi i$
(2)$\int_{|z| = 3}\frac{z}{(1-z^2)}dz=2\pi i(Res(f;1)+Res(f;-1))$
$Res(f(z);1)=\displaystyle \lim_{z\to1}\frac{(z-1)z}{(1-z^2)}=\frac {-1}2$
$Res(f(z);-1)=\displaystyle \lim_{z\to-1}\frac{(z+1)z}{(1-z^2)}=\frac {-1}2$
$\int_{|z| = 3}\frac{z}{(1-z^2)}dz=2\pi i(\frac{-1}{2}-\frac{1}{2})=-2\pi i$
