Answer
$x+6\sqrt{x}+18\ln(\sqrt{x}-3)+C$
Work Step by Step
$\int\frac{\sqrt{x}}{\sqrt{x}-3}dx$
$=\int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}dx$
$=\int1$ $dx$ $+\int\frac{3}{\sqrt{x}-3}dx$
for $\int\frac{3}{\sqrt{x}-3}dx$, let u=$\sqrt{x}$
$du=\frac{1}{2\sqrt{x}}dx$
$du=\frac{1}{2u}dx$
$2u$ $du$ $=dx$
$\int\frac{3}{\sqrt{x}-3}dx$
$=3\int\frac{2u}{u-3}du$
$=6\int\frac{u}{u-3}fu$
$=6(\int1$ $du + 3\int\frac{1}{u-3}du)$
$=6u+18ln(u-3)$
$=6\sqrt{x}+18ln(\sqrt{x}-3)$
$\int1$ $dx$ $+\int\frac{3}{\sqrt{x}-3}dx$
$=x+6\sqrt{x}+18ln(\sqrt{x}-3)+C$
