integral question

• Nov 21st 2010, 08:14 AM
Dannbr
integral question
$\displaystyle \int\frac{1}{1+\sqrt{2x}}$

according to my book it says let $\displaystyle u=1+\sqrt{2x}$

$\displaystyle du=\frac{1}{\sqrt{2x}}dx-->(u-1)du=dx$

I do not understand how $\displaystyle du=\frac{1}{\sqrt{2x}}dx-->(u-1)du=dx$

when i try to take the derivative of U i get: $\displaystyle du= \frac{d}{dx}[1+\sqrt{2x}]=\frac{2}{3}(2x)^\frac{3}{2}dx$

Im having problems getting from here:$\displaystyle \int\frac{1}{1+\sqrt{2x}}$ to here $\displaystyle \int\frac{(u-1)}{u}du$

Thanks
• Nov 21st 2010, 08:17 AM
Unknown008
$\displaystyle du=\frac{1}{\sqrt{2x}}dx$

This becomes:

$\displaystyle \sqrt{2x} du=dx$

What is $\displaystyle \sqrt{2x}$?

$\displaystyle u = 1 + \sqrt{2x}$

$\displaystyle u - 1 = \sqrt{2x}$

Hence;

$\displaystyle \sqrt{2x} du=dx$

$\displaystyle (u-1)du=dx$
• Nov 21st 2010, 08:26 AM
Unknown008
$\displaystyle du= \dfrac{d}{dx}[1+\sqrt{2x}]=\frac{1}{2}(2x)^{-\frac{1}{2}}dx$

This is what you should do. You integrated it instead of differentiating it.
• Nov 21st 2010, 08:40 AM
Dannbr
Thats it.. chain rule... got it.. thanks a lot..