Use Substitution to evaluate ${\displaystyle \underset{0}{\overset{1}{}}}(\frac{3\sqrt[\phantom{2}]{v}}{(1+v^{\frac{3}{2}}{)}^2})dv$

A$.$ Let $u=$ therefore: $du=dv$

$=>{\displaystyle \underset{0}{\overset{1}{}}}(\frac{3\sqrt[\phantom{2}]{v}}{(1+v^{\frac{3}{2}}{)}^2})dv=$

B$.$ FTC does not apply because function is not continuous at $0,1$