Answer
$\int x(3x^2+3)^3dx=\frac{1}{24}(3x^2+3)^4+C$
Work Step by Step
Substitution:
$u=3x^2+3$
$\frac{du}{dx}=6x$
$dx=\frac{1}{6x}du$
$\int x(3x^2+3)^3dx=\int xu^3\frac{1}{6x}du=\int\frac{1}{6}u^3du=\frac{1}{6}\frac{1}{4}u^4+C=\frac{1}{24}u^4+C=\frac{1}{24}(3x^2+3)^4+C$
