Answer
$\displaystyle \frac{3}{2}\ln|\sqrt{x+3}+3|+\frac{1}{2}\ln|\sqrt{x+3}-1|$
Work Step by Step
$\displaystyle \int\frac{dx}{2\sqrt{x+3}+x}=\quad\left[\begin{array}{ll}
t=\sqrt{x+3} & x=t^{2}-3\\
t^{2}=x+3 & dx=2tdt
\end{array}\right]$
$=\displaystyle \int\frac{2t}{2t+t^{2}-3}dt$
$=\displaystyle \int\frac{2t}{(t+3)(t-1)}dt$
Create partial fraction and solve for the coefficients in numerators.
$\displaystyle \frac{2t}{(t+3)(t-1)} = \frac{A}{t+3} + \frac{B}{t-1}$
$\displaystyle 2t = A(t-1) + B(t+3)$
$\displaystyle 2t = (A+B)t+(-A+3B)$
Solve the linear system.
\begin{array}{*{3}{rC}l}
A & + & B & = & 2 \\
-A & + & 3B & = & 0 \\
\end{array}
$\displaystyle A=\frac{3}{2}$, $\displaystyle B=\frac{1}{2}$
Return to solving integral, but with partial fractions.
$=\displaystyle \frac{3}{2}\int\frac{dt}{t+3} + \frac{1}{2}\int\frac{dt}{t-1}$
$=\displaystyle \frac{3}{2}\ln|t+3|+\frac{1}{2}\ln|t-1|$
$\displaystyle \frac{3}{2}\ln|\sqrt{x+3}+3|+\frac{1}{2}\ln|\sqrt{x+3}-1|$
