
integral question
$\displaystyle \int\frac{1}{1+\sqrt{2x}}$
according to my book it says let $\displaystyle u=1+\sqrt{2x}$
$\displaystyle du=\frac{1}{\sqrt{2x}}dx>(u1)du=dx$
I do not understand how $\displaystyle du=\frac{1}{\sqrt{2x}}dx>(u1)du=dx$
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 $\displaystyle \int\frac{(u1)}{u}du$
Thanks

$\displaystyle du=\frac{1}{\sqrt{2x}}dx$
This becomes:
$\displaystyle \sqrt{2x} du=dx$
What is $\displaystyle \sqrt{2x}$?
$\displaystyle u = 1 + \sqrt{2x}$
$\displaystyle u  1 = \sqrt{2x}$
Hence;
$\displaystyle \sqrt{2x} du=dx$
$\displaystyle (u1)du=dx$

$\displaystyle du= \dfrac{d}{dx}[1+\sqrt{2x}]=\frac{1}{2}(2x)^{\frac{1}{2}}dx$
This is what you should do. You integrated it instead of differentiating it.

Thats it.. chain rule... got it.. thanks a lot..