1. ## hard indefinite integral

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 . ]

2. Originally Posted by tbenne3
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 . ]
Set:
$\displaystyle u = x^3 + 1$

Note that

$\displaystyle x^5 = (u-1) x^2$

and that

$\displaystyle du = 3x^2 dx$

This should help solve for the antiderivative. Good luck!

3. Originally Posted by apcalculus
Set:
$\displaystyle u = x^3 + 1$

Note that

$\displaystyle x^5 = (u-1) x^2$

and that

$\displaystyle du = 3x^2 dx$

This should help solve for the antiderivative. Good luck!
still cant get it

4. Originally Posted by tbenne3
still cant get it
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.