Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$

$$\begin{align}\int \dfrac{6x}{\sqrt{x^2 + 4x + 8}}\, dx\;
& = \;\int \dfrac{6x+12 -12}{\sqrt{x^2 + 4x + 8}} \,dx \\ \\ \\ \\
& = \int \dfrac {6(x + 2) - 12}{\sqrt {\color{blue}{\bf (x^2+ 4x + 4) + 4}}}\,dx 
\end{align}$$
split the integral into the difference of two integrals:
$$ = \;\int \dfrac {6(x + 2)}{\sqrt {(x^2 + 4x + 8}}\,dx\;\; -\;\; 12\int \dfrac{1}{\sqrt{\color{blue}{\bf (x+2)^2 + 2^2}}}\, dx $$
Now, for the left hand integral, 


*

*let $u = x^2 + 4x + 8 \implies du = 2x + 4 = 2(x + 2) \,dx,\implies\; 3du = 6(x+2)\,dx$. Substitute: you'll have an integral of the form $\;3\int u^{-1/2} \,du$


For the right-hand integral, 


*

*let $\;v = x+2\;\implies\;dv = dx.$

*And as suggested, you'll want to use an appropriate trigonometric
substitution for the right-hand integral: in particular, find the
promising hyperbolic trig substitution.
