Pretty confused on how to do this one
Evaluate the indefinite integral.
Integral =
[NOTE: Remember to enter all necessary *, (, and ) !!
Enter arctan(x) for , sin(x) for . ]
as clearly stated earlier ...
$\displaystyle u = x^3+1$
$\displaystyle du = 3x^2 \, dx$
$\displaystyle x^3 = u - 1$
$\displaystyle \int x^5 \sqrt[3]{x^3+1} \, dx$
$\displaystyle \frac{1}{3} \int x^3 \sqrt[3]{x^3+1} \cdot 3x^2 \, dx$
substitute ...
$\displaystyle \frac{1}{3} \int (u-1) \sqrt[3]{u} \,\, du$
finish it.