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