<s>[INST] Evaluate $\int\sqrt{4x^2+8x}dx$ using $\int\sqrt{u^2-a^2}dx=\frac u2\sqrt{u^2-a^2}-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|+C.$ I completed the square to get $$2 \sqrt{(x+1)^2 - 1}$$
set  $u=x+1$ and $a=1$, and you end up with
$$(x+1) \sqrt{(x+1)^2 - 1} - \ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert+C$$
This is not the right answer. I don't see how you could simplify it more, but even if you could it would still be wrong (I put this and the answer given by Symbolab.com into Desmos.com and they didn't line up).
 [/INST] Your answer is equivalent to the answer provided by Symoblab
$$x\sqrt{x^2+2x}-\ln\left\lvert x + \sqrt{x^2+2x}+1\right\rvert+\sqrt{x^2+2x}+C$$
since starting from
$$(x+1) \sqrt{(x+1)^2 - 1} - \ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert+C$$
we have that
$$\sqrt{(x+1)^2 - 1}=\sqrt{x^2+2x}$$
therefore
$$(x+1) \sqrt{(x+1)^2 - 1}=x\sqrt{x^2+2x}+\sqrt{x^2+2x}$$
and
$$\ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert=\ln\left\lvert x + \sqrt{x^2+2x}+1\right\rvert$$
</s>