Putting $\mathrm{x^{3/2}=t~~~~i.e.~~~x=t^{2/3}}$

$\mathrm{\therefore dx=\dfrac{2}{3}t^{-1/3}~dt}$

$\mathrm{\therefore \displaystyle\int x\cdot J_{2/3}(x^{3/2})~dx=\displaystyle\int t^{2/3}\cdot J_{2/3}(t)\cdot \dfrac{2}{3}t^{-1/3}~dt}$

$\mathrm{\therefore \displaystyle\int x\cdot J_{2/3}(x^{3/2})~dx=\dfrac{2}{3}\displaystyle\int t^{1/3}~ J_{2/3}(t)~dt}$

But $\mathrm{\displaystyle\int x^{-n}\cdot J_{n+1}(x)~dx=-x^{-n} J_n(x)}$

Putting $\mathrm{n+1=\dfrac{2}{3}~~~i.e.~~~n=-\dfrac{1}{3}}$, we get

$\mathrm{\displaystyle\int x\cdot J_{2/3}(x^{3/2})~dx=\dfrac{2}{3}\bigg[-t^{-1/3}\cdot J_{-1/3}(t)\bigg]}$

$\mathrm{=-\dfrac{2}{3}\cdot (x^{3/2})^{-1/3}\cdot J_{-1/3}(x^{3/2})}$

$\mathrm{=-\dfrac{2}{3}\cdot x^{-1/2}\cdot J_{-1/3}(x^{3/2})}$

$\mathrm{\therefore \displaystyle\int x~ J_{2/3}(x^{3/2})~dx=-\dfrac{2}{3}~ x^{-1/2}~ J_{-1/3}(x^{3/2})}$

Hence proved.