# Simple Integral Help

1. Sep 29, 2010

### ebk11

1. The problem statement, all variables and given/known data

Evaluate the following integral:

[integral](x^2)*ln(x)dx

2. Relevant equations

I believe this is substitution by parts...

3. The attempt at a solution

I chose u = ln(x), and dv = x^2. The problem I am having is I can't figure out what du and v should be, because I am thinking too much and confusing myself. If u = ln(x), then du should simply be 1/x, right? And if dv = x^2, then to get v, you just need the integral of x^2, which is just 2x. Is that right?

If those are right, then how come I come up with the following answer, which I don't believe is right (and I will laugh if it is but I really don't think so):

2x*lnx - 2x + C

Last edited: Sep 29, 2010
2. Sep 29, 2010

### Dick

The INTEGRAL of x^2 isn't 2x.

3. Sep 29, 2010

### fss

No. You have differentiated, not integrated.

4. Sep 29, 2010

### ebk11

Well that's what I was confused about.

When you are setting up a substitution by parts problem...when you get u, you differentiate to get the du...and then you integrate to get v from dv, right?

So it should be x^3/3, not 2x?

5. Sep 29, 2010

### fss

Correct. Technically speaking you'd have:

u = ln x
dv = x2 dx

6. Sep 29, 2010

### ebk11

Okay, so I got du = 1/xdx, and v = x^3/3

Then, I did: lnx * x^3/3 - [integral]x^3/3 * 1/xdx

then: lnx * 1/3*x^3 - 1/3[integral]x^3/xdx

which simplifies to: lnx * 1/3 * x^3 - 1/3[integral]x^2dx

which is: lnx * 1/3 * x^3 - 1/3[integral]x^3/3 + C

final answer: ln(x) * 1/3 * x^3 - 1/9 * x^3 + C

Does that seem right?

Thanks in advance for reading this all over...

7. Sep 29, 2010

### fss

Looks like you got it to me.