Integrating x^3 lnx dx - am i on the right track?

[drjohnsonn]

Homework Statement

integrating x^3lnxdx

Homework Equations

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!

 

[scurty]

Homework Statement

integrating x^3lnxdx

Homework Equations

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

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!

 

[SammyS]

Hello drjohnsonn. Welcome to PF !

As scurty suggested, try integration by parts.

 

[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?

 

[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##

 

[SammyS]

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?

Correct.

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

 

[drjohnsonn]

Thanks for the advice!

 

[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

 

[Bohrok]

Looks good

 

[D H]

if i skipped something, point it out.

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.

 

[drjohnsonn]

wootwoot

 

[dimension10]

Homework Statement

integrating x^3lnxdx

Homework Equations

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!

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