# Simple integration problem

• Jun 3rd 2010, 12:39 PM
Positronic
Simple integration problem
I'm having trouble integrating the following:

$
\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 PM
Quote:

Originally Posted by Positronic
I'm having trouble integrating the following:

$
\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.

$
\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 $t = \sqrt{x}$.

End up with $\int \frac{2t}{t^2 - t}dt$ 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 PM
harish21
Quote:

Originally Posted by Positronic
I'm having trouble integrating the following:

$
\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.

$\int \frac{1}{x-\sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}$

now let $u = \sqrt{x}$

then $du = \frac{1}{2\sqrt{x}} dx$

and $dx = 2\sqrt{x} du = 2u \mbox{du}$

and thus you have:

$\int \frac{2u}{u(u-1)}du = \int \frac{2}{u-1}du$
.
.
.

finish it now