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!

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- Dec 15th 2008, 10:21 PMxclo0siveIndefinite 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 PMbadgerigarQuote:

$\displaystyle \int x^5\sqrt{x^3+1}dx$

- Dec 15th 2008, 10:49 PMxclo0sive
- Dec 15th 2008, 11:06 PMChris 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 $\displaystyle \dots$?? - Dec 15th 2008, 11:10 PMxclo0sive
- Dec 15th 2008, 11:13 PMbadgerigar
The book's answer is correct. I dunno what you are doing wrong though.

- Dec 15th 2008, 11:15 PMChris L T521
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 PMProve It
- Dec 15th 2008, 11:19 PMxclo0sive
- Dec 16th 2008, 12:44 AMbadgerigarQuote:

How did you get u^ 3/2

$\displaystyle =u^{1+1/2}-u^{1/2}$