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.

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- Jun 3rd 2010, 12:39 PMPositronicSimple integration problem
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.

- Jun 3rd 2010, 12:57 PMDeadstar
$\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. - Jun 3rd 2010, 01:51 PMharish21
$\displaystyle \int \frac{1}{x-\sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}$

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