# Integration - Algebraic substitution

• Jul 31st 2007, 05:39 AM
Integration - Algebraic substitution
$\displaystyle \int \frac{dx}{\sqrt{2x} - \sqrt {x+4}}$

what will i substitute?

let z^2 = x + 4
x = z^2 - 4
dx = 2zdz

but its no use
i will get a result of

$\displaystyle \int \frac{2zdz}{\sqrt{z^2-4}-z}$
need help pls
• Jul 31st 2007, 05:52 AM
topsquark
Quote:

$\displaystyle \int \frac{dx}{\sqrt{2x} - \sqrt {x+4}}$

what will i substitute?

let z^2 = x + 4
x = z^2 - 4
dx = 2zdz

but its no use
i will get a result of

$\displaystyle \int \frac{2zdz}{\sqrt{z^2-4}-z}$
need help pls

Start by rationalizing the denominator of the integrand:
$\displaystyle \int \frac{dx}{\sqrt{2x} - \sqrt {x+4}} = \int \frac{dx}{\sqrt{2x} - \sqrt {x+4}} \cdot \frac{\sqrt{2x} + \sqrt{x + 4}}{\sqrt{2x} + \sqrt{x + 4}} = \int \frac{\sqrt{2x} + \sqrt{x + 4}}{2x - (x + 4)} dx$

$\displaystyle = \int \frac{\sqrt{2x} + \sqrt{x + 4}}{x - 4} dx$

Now split the integrand into two terms and see what you can do with it.

-Dan
• Jul 31st 2007, 05:58 AM