Answer
$\frac{1}{2}[\ln|\sqrt {x+3}-1|+3\ln(\sqrt {x+3}+3)]+C$
Work Step by Step
$\int\frac{dx}{2\sqrt {x+3}+x}dx$
Substitute:
$u$ = $\sqrt {x+3}$
$u^2$ = $x+3$
$2udu$ = $dx$
$\int\frac{dx}{2\sqrt {x+3}+x}$ = $\int\frac{2u}{u^2+2u-3}du$ = $\int\frac{2u}{(u+3)(u-1)}du$ = $\int\left(\frac{A}{u-1}+\frac{B}{u+3}\right)du$
$2u$ = $A(u+3)+B(u-1)$
$0$ = $A+B$
$2$ = $3A-B$
$A$ = $\frac{1}{2}$, $B$ = $\frac{3}{2}$
So
$\int\frac{2u}{(u+3)(u-1)}du$ = $\frac{1}{2}\left(\frac{1}{u-1}+\frac{3}{u+3}\right)$ = $\frac{1}{2}[\ln|u-1|+3\ln|u+3|]+C$ = $\frac{1}{2}[\ln|\sqrt {x+3}-1|+3\ln(\sqrt {x+3}+3)]+C$
