# Antiderivatives Problem

• Nov 24th 2009, 11:36 PM
cokeclassic
Antiderivatives Problem
Hi, I am trying to find the antiderivative of this problem, but it is not the correct answer.

1/3x^2 dx

I go from that to

3x^-1/-1
what am I suppose to do with this problem?

I had a question when dividing with /x
so if a problem is 3x^2/x do simply make it 3x^1 is that correct?
And for fractions with roots at the bottom, how do I go to solving that? ex: -1/sqrt(y)

Thanks for any help, especially for the problem at the top.
Coke
• Nov 25th 2009, 01:11 AM
Andrew007
Hey there

$\displaystyle \int (x^2)/3 .dx = (x^3)/9 + C$

Yes, $\displaystyle 3x^2/x = 3x$

Not sure what your last question means..?
• Nov 25th 2009, 01:37 AM
HallsofIvy
Quote:

Originally Posted by cokeclassic
Hi, I am trying to find the antiderivative of this problem, but it is not the correct answer.

1/3x^2 dx

I go from that to

3x^-1/-1
what am I suppose to do with this problem?

I had a question when dividing with /x
so if a problem is 3x^2/x do simply make it 3x^1 is that correct?

As long as x is not equal to 0, yes.

Quote:

And for fractions with roots at the bottom, how do I go to solving that? ex: -1/sqrt(y)
What do you mean by "solve" that? What do you want to do with it? If you just want a different way to write it, perhaps to integrate, you could try $\displaystyle -y^{1/2}$.

Quote:

Thanks for any help, especially for the problem at the top.
Coke
• Nov 25th 2009, 06:43 AM
cokeclassic
Hi, I am not sure what you did with the first problem? Could you show me some steps please?