integral question

**Dannbr** · November 21st 2010, 08:14 AM

$\int\frac{1}{1+\sqrt{2x}}$

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

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

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

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

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

Thanks

**Unknown008** · November 21st 2010, 08:17 AM

$du=\frac{1}{\sqrt{2x}}dx$

This becomes:

$\sqrt{2x} du=dx$

What is $\sqrt{2x}$ ?

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

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

Hence;

$\sqrt{2x} du=dx$

$(u-1)du=dx$

**Unknown008** · November 21st 2010, 08:26 AM

$du= \dfrac{d}{dx}[1+\sqrt{2x}]=\frac{1}{2}(2x)^{-\frac{1}{2}}dx$

This is what you should do. You integrated it instead of differentiating it.

**Dannbr** · November 21st 2010, 08:40 AM

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

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