I am stuck!

$\displaystyle \int\frac{dx}{\sqrt{1-2x^{2}}}$

$\displaystyle \mbox{Let }u=1-2x^{2}, du=-4xdx$

$\displaystyle \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{u^{-1/2}du}{x}$

$\displaystyle \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{du}{u^{1/2}(1/2-u/2)^{1/2}}$

$\displaystyle \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{du}{(u/2-u^{2}/2)^{1/2}}$

$\displaystyle \int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-\sqrt{2}}{4}\int\frac{du}{(u-u^{2})^{1/2}}$

I don't know where to go from there. Any help is greatly appreciated.