How to solve $\int \frac{1}{1-y^2}$ with respect to $y$? I was solving an A Level paper when I came across this question. I tried substitution, but I'm not getting the answer with that. Would appreciate it if someone would help me.

Using a simple substition of $y = \tanh{(u)}$ gives
$\displaystyle\int \dfrac{1}{1-y^{2}} \mathrm{d}y = u + C = \tanh^{-1}y + C = \ln{\dfrac{1 + y}{1 - y}} + C$
Notably can also be solved using partial fractions. 
