Complete the square:
\(\displaystyle\int\sqrt{x^2+6x}\,\mathrm dx=\int\sqrt{(x+3)^2-9}\,\mathrm dx\)
Now substitute \(x+3=3\sec y\), so that \(\mathrm dx=3\sec y\tan y\,\mathrm dy\). The integral then becomes
\(\displaystyle\int\sqrt{(3\sec y)^2-9}(3\sec y\tan y)\,\mathrm dy\)
\(\displaystyle9\int\sqrt{\sec^2y-1}\sec y\tan y\,\mathrm dy\)
\(\displaystyle9\int\sqrt{\tan^2y}\sec y\tan y\,\mathrm dy\)
\(\displaystyle9\int\sec y\tan^2y\,\mathrm dy\)
Use the Pythagorean theorem to reduce the integrand to
\(\displaystyle9\int\sec y(\sec^2y-1)\,\mathrm dy\)
\(\displaystyle9\int(\sec^3y-\sec y)\,\mathrm dy\)
You can integrate \(\sec^3y\) by parts, setting
\(\begin{matrix}u=\sec y&&\mathrm dv=\sec^2y\,\mathrm dy\\\mathrm du=\sec y\tan y\,\mathrm dy&&v=\tan y\end{matrix}\)
So,
\(\displaystyle\int\sec^3y\,\mathrm dy=\sec y\tan y-\int\sec y\tan^2y\,\mathrm dy\)
\(\displaystyle2\int\sec^3y\,\mathrm dy=\sec y\tan y+\int\sec y\,\mathrm dy\)
\(\displaystyle\int\sec^3y\,\mathrm dy=\frac12\sec y\tan y+\frac12\int\sec y\,\mathrm dy\)
This means you have
\(\displaystyle9\left(\frac12\sec y\tan y+\frac12\int\sec y\,\mathrm dy\right)-9\int\sec y\,\mathrm dy\)
\(\displaystyle\frac92\sec y\tan y-\frac92\int\sec y\,\mathrm dy\)
You can integrate \(\sec y\) by writing
\(\displaystyle\int\sec y\frac{\sec y+\tan y}{\sec y+\tan y}\,\mathrm dy=\int\frac{\sec^2y+\sec y\tan y}{\sec y+\tan y}\,\mathrm dy=\int\frac{\mathrm dz}z=\ln|\sec y+\tan y|+C\)
So you are left with
\(\displaystyle\frac92\sec y\tan y-\frac92\ln|\sec y+\tan y|+C\)
Transforming back to \(x\) gives you
\(\displaystyle\frac92\sec\left(\arcsec\frac{x+3}3\right)\tan\left(\arcsec\frac{x+3}3\right)-\frac92\ln\left|\sec\left(\arcsec\frac{x+3}3\right)+\tan\left(\arcsec\frac{x+3}3\right)\right|+C\)
\(\displaystyle\frac92\frac{x+3}3\frac{\sqrt{x^2+6x}}3-\frac92\ln\left|\frac{x+3}3+\frac{\sqrt{x^2+6x}}3\right|+C\)
\(\displaystyle\frac{(x+3)\sqrt{x^2+6x}}2-\frac92\ln\left|\frac{x+3+\sqrt{x^2+6x}}3\right|+C\)
\(\displaystyle\frac{(x+3)\sqrt{x^2+6x}}2-\frac92\ln\left|x+3+\sqrt{x^2+6x}\right|+C\)
