## Calculus: Early Transcendentals 8th Edition

all real $u$, $u\ne-1,-2$
$\because$ the denominator of $\frac{1}{u+1}\ne0,$ $u+1\ne0, u\ne-1$ $f(u)=\frac{(u+1)(u+1)}{(u+1)+1} =\frac{u^{2}+2u+1}{u+2}$ $\because$ the denominator of $f(u)\ne0$, $u+2\ne0, u\ne-2$ $\therefore$ the domain is all real $u$, excluding $u=-1,-2$