How to compute the integral $ \int\frac{1}{x\sqrt{x^2 +3x}}dx$ Given a problem :
$$ \int\frac{1}{x\sqrt{x^2 +3x}}dx, $$
what is the best solution for this?
I am thinking about solving this problem by using :
$$ u = x+3;\qquad x = u-3; $$
So that we get :
$ \int\frac{1}{x\sqrt{x}\sqrt{x+3}} dx$, then $ \int\frac{1}{(u-3)^{3/2}(u)^{1/2}} du$,
then 
$ \int (u)^{-1/2} (u-3)^{-3/2}  du$.
Am I right so far? or is there a better method? Thanks.

For $x>0$, $$\int\frac{1}{x\sqrt{x^2 +3x}}dx=\int\frac{1}{x^2\sqrt{1 +\frac{3}{x}}}dx = -\frac{1}{3}\int\frac{du}{\sqrt{u}}=-\frac{2}{3}\sqrt{u}=-\frac{2}{3}\sqrt{1+\frac{3}{x}} +C $$ where $$u=1+\frac{3}{x}$$
