indefinate integral

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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?

Please help?
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?

Please help?

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

note that $\frac 2{3x^{1/3}} = \frac 23x^{-1/3}$ . now integrate that using the power rule for integrals $\left( \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C \text{ for } n \ne -1 \right)$
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?

please post one problem per thread.

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