# Indefinite integral by compound substitution

• Dec 15th 2008, 10:21 PM
xclo0sive
Indefinite integral by compound substitution
Need help!

Integral of (X^5) sqrt (x^3)+1 dx

I hope this isint too hard to read but its only the sqrt of x^3+1
Thanks!
• Dec 15th 2008, 10:35 PM
Quote:

$\displaystyle \int x^5\sqrt{x^3+1}dx$
try substituting $\displaystyle u = x^3+1$
• Dec 15th 2008, 10:49 PM
xclo0sive
Quote:

Originally Posted by badgerigar
try substituting $\displaystyle u = x^3+1$

I tried that, even split x^5 to (x^2)(x^3) but my answer is way off from the books. btw the answer is 2/15((x^3)+1)^5/2 - 2/9((x^3)+1)^3/2+c
• Dec 15th 2008, 11:06 PM
Chris L T521
Quote:

Originally Posted by xclo0sive
I tried that, even split x^5 to (x^2)(x^3) but my answer is way off from the books. btw the answer is 2/15((x^3)+1)^5/2 - 2/9((x^3)+1)^3/2+c

Going along with that u-sub, you get $\displaystyle \,du=3x^2\,dx$

Thus, you now get $\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du$

But $\displaystyle u=x^3+1\implies\dots$

So then you should end up with $\displaystyle \tfrac{1}{3}\int\dots\,du$

Can you fill in the $\displaystyle \dots$??
• Dec 15th 2008, 11:10 PM
xclo0sive
Quote:

Originally Posted by Chris L T521
Going along with that u-sub, you get $\displaystyle \,du=3x^2\,dx$

Thus, you now get $\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du$

But $\displaystyle u=x^3+1\implies\dots$

So then you should end up with $\displaystyle \tfrac{1}{3}\int\dots\,du$

Can you fill in the $\displaystyle \dots$??

my final answer was 1/12 (x 2/3((x^3)+1)^3/2+c but its nowhere close to the answer provided.
• Dec 15th 2008, 11:13 PM
The book's answer is correct. I dunno what you are doing wrong though.
• Dec 15th 2008, 11:15 PM
Chris L T521
Quote:

Originally Posted by xclo0sive
my final answer was 1/12 (x 2/3((x^3)+1)^3/2+c but its nowhere close to the answer provided.

Quote:

Originally Posted by Chris L T521
Going along with that u-sub, you get $\displaystyle \,du=3x^2\,dx$

Thus, you now get $\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du$

But $\displaystyle u=x^3+1\implies\dots$

So then you should end up with $\displaystyle \tfrac{1}{3}\int\dots\,du$

Can you fill in the $\displaystyle \dots$??

Since $\displaystyle u=x^3+1$, it implies that $\displaystyle u-1=x^3$

Thus, $\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du=\tfrac{1}{3}\int\left(u-1\right)\sqrt{u}\,du=\tfrac{1}{3}\int \left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right)\,du$

Can you take it from here?
• Dec 15th 2008, 11:16 PM
Prove It
Quote:

Originally Posted by Chris L T521
Going along with that u-sub, you get $\displaystyle \,du=3x^2\,dx$

Thus, you now get $\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du$

But $\displaystyle u=x^3+1\implies\dots$

So then you should end up with $\displaystyle \tfrac{1}{3}\int\dots\,du$

Can you fill in the $\displaystyle \dots$??

$\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du$

$\displaystyle u=x^3+1\implies\ x^3 = u - 1$

So $\displaystyle \tfrac{1}{3}\int (u - 1)\sqrt{u}\,du = \tfrac{1}{3}\int u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du$.

Can you do this integral?
• Dec 15th 2008, 11:19 PM
xclo0sive
Quote:

Originally Posted by Prove It
$\displaystyle \tfrac{1}{3}\int x^3\sqrt{u}\,du$

$\displaystyle u=x^3+1\implies\ x^3 = u - 1$

So $\displaystyle \tfrac{1}{3}\int (u - 1)\sqrt{u}\,du = \tfrac{1}{3}\int u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du$.

Can you do this integral?

Oh WOW THanks guys! I completely forgot to substitute the x^3! Thanks!!
EDIT: How did you get u^ 3/2
• Dec 16th 2008, 12:44 AM
$\displaystyle (u-1)\sqrt{u} = u^1u^{1/2}-u^{1/2}$
$\displaystyle =u^{1+1/2}-u^{1/2}$