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.

Hint: Let $y^2= 1+\dfrac{3}{x}$, i.e. $x=\dfrac{3}{y^2-1}$.

 Then $dx=-\dfrac{6ydy}{(y^2-1)^2}$.

