Answer
$$\frac{(x+3) \sqrt{x^{2}+6 x}}{2}-\frac{9}{2} \ln \left|\frac{(x+3)+\sqrt{x^{2}+6 x}}{3}\right|$$
Work Step by Step
Given $$\int \sqrt{x^{2}+6 x} d x$$
Since
$$\int \sqrt{x^{2}+6 x} d x =\int \sqrt{(x+3)^{2}-9} d x$$
Let $$ x+3=3\sec u,\ \ \ \ \ \ \ dx=3\sec u\tan udu$$
Then
\begin{align*}
\int \sqrt{x^{2}+6 x} d x&=3\int \sqrt{9\sec^2u-9}\sec u\tan udu \\
&=9\int \tan^2u \sec udu\\
&=9\int( \sec^3 u-\sec u )du,\ \\
& \text{From the table: }\\
&=9\left(\frac{1}{2}\sec \left(u\right)\tan \left(u\right)-\frac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|+C\right)\\
&= \frac{(x+3) \sqrt{x^{2}+6 x}}{2}-\frac{9}{2} \ln \left|\frac{(x+3)+\sqrt{x^{2}+6 x}}{3}\right|
\end{align*}
