Answer
$-\frac{1}{2(x^2+2x-3)}+C$
Work Step by Step
$\frac{dy}{dx}=\frac{x+1}{(x^2+2x-3)^2}$
$\int\frac{dy}{dx}dx=\int\frac{x+1}{(x^2+2x-3)^2}dx$
Let $u=x^2+2x-3\hspace{8mm}du=(2x+2)dx=2(x+1)dx$
$dx=\frac{du}{2(x+1)}$
$\int\frac{x+1}{(x^2+2x-3)^2}dx=\int\frac{x+1}{(u)^2}\frac{du}{2(x+1)}=\frac{1}{2}\int\frac{1}{u^2}du$
$=\frac{1}{2}\int u^{-2}du=\frac{1}{2}(-u^{-1})+C=-\frac{1}{2u}+C$
$=-\frac{1}{2(x^2+2x-3)}+C$
