Answer
The equation has two solutions, 0 and 2.
Work Step by Step
$\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{u^2-u-6}$
$\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{(u+2)(u-3)}$
$(u+2)(u-3)\frac{u+1}{u+2}=(u+2)(u-3)(\frac{-1}{u-3}+\frac{u-1}{(u+2)(u-3)})$
$(u+1)(u-3)=-1(u+2)+u-1$
$u^2-2u-3=-u-2+u-1$
$u^2-2u-3=-3$
$u^2-2u=0$
$u(u-2)=0$
$u=0$ or $u-2=0$
$u=0$ or $u=2$
Check:
$\frac{0+1}{0+2}=1/2; \frac{-1}{0-3}+\frac{0-1}{0-0-6}=1/3+1/6=1/2$
$\frac{2+1}{2+2}=3/4; \frac{-1}{2-3}+\frac{2-1}{4-2-6}=1+(-1/4)=3/4$
The equation has two solutions, 0 and 2.
