Answer
$\frac{x^{2}+x+4}{(x-1)(x+1)^{2}}$
Work Step by Step
We form a common denominator, then add, factor, and simplify:
$\displaystyle \frac{1}{x+1}-\frac{2}{(x+1)^{2}}+\frac{3}{x^{2}-1}=\frac{1}{x+1}-\frac{2}{(x+1)^{2}}+\frac{3}{(x-1)(x+1)}
=\frac{(x+1)(x-1)}{(x-1)(x+1)^{2}}+\frac{-2(x-1)}{(x-1)(x+1)^{2}}+\frac{3(x+1)}{(x-1)(x+1)^{2}}
=\frac{x^{2}-1}{(x-1)(x+1)^{2}}+\frac{-2x+2}{(x-1)(x+1)^{2}}+\frac{3x+3}{(x-1)(x+1)^{2}}=\frac{x^{2}-1-2x+2+3x+3}{(x-1)(x+1)^{2}}=\frac{x^{2}+x+4}{(x-1)(x+1)^{2}}$
