Answer
$u=0,2$
Work Step by Step
Given:$\dfrac{u+1}{u+2}=\dfrac{-1}{u-3}+\dfrac{u-1}{(u^2-u-6)}$
Need to find least common denominator(LCD).
LCD: $(u+1)(u+2)(u^2-u-6)$
$\dfrac{u+1}{u+2}=\dfrac{-1}{u-3}+\dfrac{u-1}{(u^2-u-6)}$
$\dfrac{(u+1)(u-3)+(u+2)}{u^2-u-6}=\dfrac{u-1}{(u^2-u-6)}$
$(u+1)(u-3)+(u+2)=(u-1)$
$u^2-2u-3+u+2-u+1=0$
$u(u-2)=0$
Hence, $u=0,2$