My English-Math language isn't too good, so I'll try to use math to explain:
$\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$
$\displaystyle t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$
$\displaystyle \Downarrow$
$\displaystyle e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$
$\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$ = $\displaystyle \int \sqrt{x} t*2\sqrt{x}(1+\sqrt{x})dt$
= $\displaystyle \int 2t(x+x^\frac{3}{2}) dt$
=$\displaystyle \int 2t((e^t-1)^2+(e^t-1)^3) dt $
=$\displaystyle \int 2t*e^t(e^t-1)^2)) dt $
Now what

?
Thank you very much
