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

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