$I_1=\int\limits_0^1\dfrac {x^2}{\sqrt{1-x^4}}dx$ Put $x^4=t,x^2=t^{1/2},x=t^{1/2}$ $dx=\dfrac 14t^{-3/4}dt$ When $x=1,t=1$ $x=0,t=0$ $I_1=\int\limits_0^1\dfrac {t^{1/2}\times\dfrac 14t^{-3/4}dt}{(1-t)^{1/2}}$ $I_1=\dfrac 14\int\limits_0^1t^{-1/4}(1-t)^{-1/4}dt$ $=\dfrac 14 B\Bigg(\dfrac 34,\dfrac 12\Bigg)$ $I_2=\int\limits_0^1\dfrac 1{\sqrt{1-x^4}}dx$ Put $x^4=t , x=t^{1/4}$ $dx=\dfrac 14t^{-3/4}dt$ When $x=0,t=0$ $x=1,t=1$ $I_2=\int\limits_0^1\dfrac{\dfrac 14t^{-3/4}}{\sqrt{1-t}}$ $I_2=\dfrac14\int\limits^1_0t^{-3/4}(1-t)^{-1/2}dt$ $I_2=\dfrac 14B\Bigg(\dfrac 14,\dfrac 12\Bigg)$ $\therefore I_1\times I_2= I$ $\therefore I=\dfrac 14B\Bigg(\dfrac 34,\dfrac 12\Bigg)\times \dfrac 14B\Bigg(\dfrac 14,\dfrac 12\Bigg)$ $=\dfrac 1{16}\dfrac {\sqrt{\dfrac 34}\sqrt{\dfrac 12}}{\sqrt{\dfrac 54}}\times \dfrac {\sqrt{\dfrac 14}\sqrt{\dfrac 12}}{\sqrt{\dfrac 34}}$ $=\dfrac 1{16}\dfrac {\sqrt{\dfrac 12}^2\times\sqrt{\dfrac 14}}{\dfrac 14\sqrt{\dfrac 14}}$ $=\dfrac 14\sqrt{\pi}^2$ $=\dfrac \pi4$