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

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

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- Jul 31st 2007, 05:39 AM^_^Engineer_Adam^_^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 AMtopsquark
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^_^Engineer_Adam^_^
got it now thanks ts!