Answer
$\frac{1}{3}(3x^{4}+2)^{\frac{3}{2}}+c$
Where c is an arbitrary constant.
Work Step by Step
$\int$$6x^{3}$ $\sqrt (3x^{4}+2)$ dx
=$\frac{1}{2}\int12x^{3}(3x^{4}+2)^{\frac{1}{2}}$
=$\frac{1}{2}[\frac{(3x^{4}+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c]$
=$\frac{1}{2}\times\frac{2}{3}(3x^{4}+2)^{\frac{3}{2}}+c$
=$\frac{1}{3}(3x^{4}+2)^{\frac{3}{2}}+c$
Where c is an arbitrary constant.
