Answer
$\frac{(2t^4+3)^{\frac{3}{2}}}{12}+C$
Work Step by Step
$\int t^3 (2t^4+3)^{\frac{1}{2}}dt$
let
$2t^4+3=u$
$8t^3dt=du$
$t^3dt=\frac{1}{3}du$
$\int t^3 (2t^4+3)^{\frac{1}{2}}dt$
$=\frac{1}{8} \int u^{\frac{1}]{2}}$
$=\frac{1}{12}u^{\frac{3}{2}}+C$
$=\frac{(2t^4+3)^{\frac{3}{2}}}{12}+C$
