# indefinate integral

• May 12th 2009, 08:52 PM
craziebbygirl
indefinate integral
Hi I am a bit confused on how to differentiate a fraction like the problem:

integral of 2 / (3 x^1/3) dx

Im not sure where to start can you pull the 2/3 out front and just make it:

1/ x^1/3? and take ln of x^1/3?

• May 12th 2009, 08:58 PM
Jhevon
Quote:

Originally Posted by craziebbygirl
Hi I am a bit confused on how to differentiate a fraction like the problem:

integral of 2 / (3 x^1/3) dx

Im not sure where to start can you pull the 2/3 out front and just make it:

1/ x^1/3? and take ln of x^1/3?

the natural log only comes into play if you have 1/x, which you do not have.

note that $\displaystyle \frac 2{3x^{1/3}} = \frac 23x^{-1/3}$. now integrate that using the power rule for integrals $\displaystyle \left( \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C \text{ for } n \ne -1 \right)$
• May 12th 2009, 09:03 PM
craziebbygirl
Thanks! so would the answer be 1/2 x^4/3 + c?
• May 12th 2009, 09:06 PM
craziebbygirl
also:
I have the problem: find the integral of (5x - 6) / (square root of X) dx

Can you separate it into 5x / X^1/2 - 6/x^1/2? even if I do that I still don't see how to solve it?
• May 12th 2009, 09:20 PM
Jhevon
Quote:

Originally Posted by craziebbygirl
Thanks! so would the answer be 1/2 x^4/3 + c?

no. look at the rule again. note that you have a negative 1/3 power.

Quote:

Originally Posted by craziebbygirl
also:
I have the problem: find the integral of (5x - 6) / (square root of X) dx

Can you separate it into 5x / X^1/2 - 6/x^1/2? even if I do that I still don't see how to solve it?

Hint: $\displaystyle \frac {x^a}{x^b} = x^{a - b}$