## College Algebra 7th Edition

$\frac{2}{(x-1)(x+3)}$
We multiply by $(x-1)(x+3)$, distribute and simplify: $\displaystyle \frac{\frac{1}{x-1}+\frac{1}{x+3}}{x+1}=\frac{(x-1)(x+3)(\frac{1}{x-1}+\frac{1}{x+3})}{(x-1)(x+3)(x+1)}=\frac{(x+3)+(x-1)}{(x-1)(x+1)(x+3)}=\frac{2(x+1)}{(x-1)(x+1)(x+3)}=\frac{2}{(x-1)(x+3)}$