Not sure if i've done this problem right.

$\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

$\displaystyle u=2x^3+7$

$\displaystyle du=6x^2$

$\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

$\displaystyle \int (\frac{du}{u^{3/2}})$

$\displaystyle \int (2u^{-3/2}+c$

$\displaystyle 2(2x^3+7)^{-3/2}+c$

$\displaystyle (4x^3+14)^{-3/2}+c$