## Thomas' Calculus 13th Edition

$\frac{2}{5}(x^3+1)^{5/2}-\frac{2}{3}(x^3+1)^{3/2}+C$
Let u=$x^3+1=>du=3x^2dx$ and $x^3=u-1$ so $\int 3x^5 \sqrt{x^3+1}dx$ =$\int (u-1)\sqrt{u}$ =$\int (u^{3/2}-u^{1/2})du$ =$\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}+C$ =$\frac{2}{5}(x^3+1)^{5/2}-\frac{2}{3}(x^3+1)^{3/2}+C$