# integration

• Nov 11th 2009, 03:49 PM
Nightasylum
integration
integral
2x^2(x^3+1)^(1/2) dx
Having the hardest time with this is there a simple way to do it.
• Nov 11th 2009, 03:51 PM
Jester
Quote:

Originally Posted by Nightasylum
integral
2x^2(x^3+1)^(1/2) dx
Having the hardest time with this is there a simple way to do it.

For $\displaystyle \int 2x^2 \sqrt{x^3+1}\,dx$ let $\displaystyle u = x^3+1$. See how that works?
• Nov 11th 2009, 03:53 PM
skeeter
Quote:

Originally Posted by Nightasylum
integral
2x^2(x^3+1)^(1/2) dx
Having the hardest time with this is there a simple way to do it.

substitution ...

$\displaystyle u = x^3+1$

$\displaystyle du = 3x^2 \, dx$

$\displaystyle \frac{2}{3} \int 3x^2\sqrt{x^3+1} \, dx$

$\displaystyle \frac{2}{3} \int \sqrt{u} \, du$

finish ...
• Nov 11th 2009, 03:57 PM
Nightasylum
Quote:

Originally Posted by skeeter
substitution ...

$\displaystyle u = x^3+1$

$\displaystyle du = 3x^2 \, dx$

$\displaystyle \frac{2}{3} \int 3x^2\sqrt{x^3+1} \, dx$

$\displaystyle \frac{2}{3} \int \sqrt{u} \, du$

finish ...

Ill get 2/3 (x^3+1)^3/2 + c But that doesnt work so im a little lost.
• Nov 11th 2009, 04:03 PM
skeeter
Quote:

Originally Posted by Nightasylum
Ill get 2/3 (x^3+1)^3/2 + c But that doesnt work so im a little lost.

... you ignored the constant in front of the integral.
• Nov 11th 2009, 04:15 PM
Nightasylum
Quote:

Originally Posted by skeeter
... you ignored the constant in front of the integral.

O ok I got it. Thanks a ton .