I'm having trouble integrating the following:
$\displaystyle
\int \frac{1}{x-\sqrt{x}} dx
$
I can tell that I'm going to have to do a few substitutions... but I have no idea where.
$\displaystyle
\int \frac{1}{x-\sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx
$
Then use partial fractions... (Haven't checked if it works though)
edit: Don't think this will help actually lol.
Or alternatively
set $\displaystyle t = \sqrt{x}$.
End up with $\displaystyle \int \frac{2t}{t^2 - t}dt$ which you can simplify and will be easier to solve.
EDIT 2: Got t^2 and t mixed up. Above is correct.
$\displaystyle \int \frac{1}{x-\sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}$
now let $\displaystyle u = \sqrt{x}$
then $\displaystyle du = \frac{1}{2\sqrt{x}} dx$
and $\displaystyle dx = 2\sqrt{x} du = 2u \mbox{du}$
and thus you have:
$\displaystyle \int \frac{2u}{u(u-1)}du = \int \frac{2}{u-1}du$
.
.
.
finish it now