Question

${\int }_0^{2\pi }$ $\frac{\sqrt{1+x^{1/3}}}{x^{2/3}}$ dx

Answer 1

${\int }_0^{2\pi }\frac{\sqrt{1+x^{1/3}}}{x^{2/3}}dx$

Substitute $u=1+x^{1/3}\to dx=3x^{2/3}du$

$=3{\int }_0^{2\pi }\sqrt{u}du$

$=3{\left[\frac{2u^{3/2}}{3}\right]}_0^{2\pi }$

$=[2u^{3/2}{]}_0^{2\pi }$

$=\left[2\right(1+x^{1/3}{)}^{3/2}{]}_0^{2\pi }$

$=\left[2\right(1+2{\pi }^{1/3}{)}^{3/2}]-[2(1+0{)}^{3/2}]$

$=2\left[\right(1+2{\pi }^{1/3}{)}^{3/2}-1]$.