# Homework Help: Integrating x^3 lnx dx - am i on the right track?

1. May 27, 2012

### drjohnsonn

1. The problem statement, all variables and given/known data
integrating x^3lnxdx

2. Relevant equations

3. The attempt at a solution

i let u = lnx

du/dx = 1/x
xdu = dx
x=e^u
substituting that, i got e^(4u)udu

then i let v = 4u
dv/du = 4
1/4du = dv

substituting that, i got 1/4integral e^v vdv

I haven't gone beyond that step yet. I was hoping someone more knowledgeable than I currently am could tell me if I am on the right track or let me know at what point I went wrong. Responses very much appreciated!

Last edited: May 27, 2012
2. May 27, 2012

### scurty

You're fine up to this point! Using u-substitution won't do anything helpful for you at this point; you are basically left trying to integrate the same problem. I would suggest integration by parts!

3. May 27, 2012

### SammyS

Staff Emeritus
Hello drjohnsonn. Welcome to PF !

As scurty suggested, try integration by parts.

4. May 27, 2012

### drjohnsonn

Thanks! I am quite a novice as of now and have not quite gotten to that point yet. I suppose I'll have to wait a bit before I finish working on this. But is the reason for having to integrate by parts that there is no general product integration rule?

5. May 27, 2012

### scurty

Well, I immediately think integration by parts when I see a term that can be reduced by differentiating (u in this case) and a term that doesn't get "more complicated" by anti-differentiating ($e^{4u}$ in this case). It's tough to explain, someone else might have a better general rule for when to integrate by parts.

Also, for what it's worth, you could have integrated by parts right from the start with no u-substitution. $u = ln(x), \quad dv = x^3 \ dx$

6. May 27, 2012

### SammyS

Staff Emeritus
Correct.

In fact integration by parts is based upon the product rule for differentiation.

7. May 28, 2012

### drjohnsonn

8. May 28, 2012

### drjohnsonn

Am I accurate in doing this:
u = lnx dv=x^3dx
du = 1/xdx v = Sx^3dx = x^4/4

and since Sudv = uv - Svdu

Sx^3lnxdx = 1/4(x^4)lnx - 1/4Sx^3dx = 1/4x^4lnx - 1/16x^4 + C

if i skipped something, point it out. i kind of ended up looking at the answer before i finished and hope that didn't fool me into incorrectly solving it

9. May 28, 2012

### Bohrok

Looks good

10. May 28, 2012

### D H

Staff Emeritus
There's an easy way to check: Differentiate. Differentiation is oftentimes far easier than integration (and the only time it isn't is when both are easy). Upon differentiation you should recover the original equation.

It's always a good idea to double check your work.

11. May 28, 2012

### drjohnsonn

wootwoot

12. May 28, 2012

### dimension10

Yes. I think you are on the right track. Now, you can do integration by parts again to get the answer.