Integration of powers of natural logs

[sciencefog]

Hi,

I need some help! :uhh:

I have a series of maths problems in my physics work...

I am trying to integrate a power of a log - such as:

y= 3.2875E+004 - 8.9321E+005*ln x + 4.2316E+006*(ln x)^2 - 7.7776E+006*(ln x)^3 + 7.4949E+006*(ln x)^4 - 4.1373E+006*(ln x)^5 + 1.3164E+006*(ln x)^6 - 2.2448E+005*(ln x)^7 + 1.5873E+004*(ln x)^8

Any help at all would be great :smile:

All I have is that: integral of lnx = xlnx - x + C

even just the simple y = 2*(lnx)^3, would be great, I can do the rest from there...I assume its a 'by parts' expression?

Cheers :redface:

[Hurkyl]

I assume its a 'by parts' expression?

Sounds reasonable; have you tried it?

[sciencefog]

Well I was thinking - but I don't know if its right and I don't know where to find out - hence posting here... :smile:

y = 2*(lnx)^3

= 2*[(lnx)(lnx)(lnx)]

therefore integral = 2*( xlnx - x * xlnx - x * xlnx - x) +C
= 2*(3xlnx - x) + c
=6xlnx - x + c

Is that right?

Thanks very much for the help. :biggrin:

[arildno]

This is absolutely wrong!!
(Ask yourself: How can I prove that my answer is wrong? Hint: Differentiate..)

[sciencefog]

ok, so can you help me get further than this? maybe pointing me in the direction of the parts bit?

I know that: (all this is from the web, I have done this work - but 6 years ago..)

integral = $

$udv = uv - $vdu

[eg $xe^x.dx, u = x, dv = e^x.dx]

so for the expression: $2(lnx)^3.dx

I need a u part and a dv part:

u = lnx, (dv = x^3.dx)??? I'm not sure what the dv part should be??

I also assume I take the constant out?

Cheers :redface:

[arildno]

Set u=(ln(x))^3 v'=1
Then,
uv=xln(x)^3, vu'=3ln(x)^2
Then, use the same technique with (ln(x))^2

[sciencefog]

Why is dv/dx = 1?

so here is my attempt at an answer:

$2(lnx)^3

u=(lnx)^3, v'=1

uv = x(lnx)^3, vu'=3(lnx)^2

u = (lnx)^2, v'=1

uv= x(lnx)^2, vu'=2lnx

so: $udv=x(lnx)^3 -$3(lnx)^2
= x(lnx)^3 - x(lnx)^2 - $2lnx
=x(lnx)^3 - x(lnx)^2 - 2(xlnx - x)

so complete answer would be:
= 2(x(lnx)^3 - x(lnx)^2 - 2(xlnx - x)) + C ??

[sciencefog]

I really am confused with the v' =1 i.e. are you saying that the integral of lnx = 1?? :cry:

How can that be - as when you differentiate you do not get lnx???

If you are not saying that then what is going on = can you break it down for me please as I'm very confused...

I thought the integral of lnx = xlnx - x +C ????

also - if you have: $2(lnx)^3 how is taking u = (lnx)^3 breaking it down - isn't that the same as what you start with - ie needing to break into parts...

I'm sorry if this is really obvious - but I am having real trouble working this out...

Cheers

[arildno]

You are mixing up dv and v'!
The integration by parts formula reads:
Int(uv',dx)=uv-Int(u'v,dx)
You may write this (using the change of variables formula) as:
Int(u,dv)=uv-Int(v,du)

[reptile101]

dats helpd a lil man... i was kinda stuck on da integral of (lnx)^2, an i thought dat mayb cos yu got (lnx) to the power 2, dat yu cnt take u = (lnx)^2... but now im guessin yu can... RIIGHT? :uhh: :confused: