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$\int\limits_{0}^{1/4}$ $\sqrt{\frac{x}{1-2x}}$dx Đặt $t =\sqrt{x}=> t^2=x=>2tdt=dx$x 0 1/4t 0 1/2= $\int\limits_{0}^{1/2}\frac{t.2t.dt}{\sqrt{1-2t^2}}$Đặt $t=\frac{1}{\sqrt{2}}sinu=>dt=\frac{1}{\sqrt{2}}cosu.du$t 0 1/2u 0 $\Pi/4$= $\int\limits_{0}^{\Pi/4}\frac{2(\frac{1}{\sqrt{2}}sinu)^2\frac{1}{\sqrt{2}}cosu.du}{\sqrt{1-sin^2u}}$= $\int\limits_{0}^{\Pi/4}sin^2u.\frac{1}{\sqrt{2}}.du$

$\int\limits_{0}^{1/4}$ $\sqrt{\frac{x}{1-2x}}$dx Đặt $t =\sqrt{x}=> t^2=x=>2tdt=dx$x 0 1/4t 0 1/2= $\int\limits_{0}^{1/2}\frac{t.2t.dt}{\sqrt{1-2t^2}}$Đặt $t=\frac{1}{\sqrt{2}}sinu=>dt=\frac{1}{\sqrt{2}}cosu.du$t 0 1/2u 0 $\Pi/4$= $\int\limits_{0}^{\Pi/4}\frac{2(\frac{1}{\sqrt{2}}sinu)^2\frac{1}{\sqrt{2}}cosu.du}{\sqrt{1-sin^2u}}$= $\int\limits_{0}^{\Pi/4}sin^2u.\frac{1}{\sqrt{2}}.du$= $\frac{1}{\sqrt{2}}\int\limits_{0}^{\Pi/4}\frac{1-cos2x}{2}.du$=$\frac{1}{2\sqrt{2}}(u-\frac{1}{2}sin2u)|\begin{matrix} \Pi/4\\ 0 \end{matrix}$ =$\frac{\Pi-2}{8\sqrt{2}}$