Mathematical Analysis I - 4-4-4.(2)

$\int_0^1 \frac{\sqrt{x}}{1+\sqrt[3]{x}} dx$ Take $x = t^6$ , the integral becomes $6 \int_0^1 \frac{t^8}{1+t^2}dt$ $=6\int_0^1 (t^4+1)(t^2 -1) + \int_0^1 \frac{1}{1+t^2} dt$ $= 6\int_0^1 (t^6+t^2 -t^4 -1)dt + 6[ \tan^{-1}t]_0^{1}$ $= -\frac{152}{35} +\frac{3π}{2}$ $= \frac{105π-304}{70}$ $=\boxed{0.369531}$ Answer is $⌊10000×0.369531⌋ =\color{#20A900}\boxed{3695}$