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