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Integrating 1/xln(x) using integration by parts

by ductruffe
Tags: integration parts ln
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ductruffe
#1
Sep13-10, 08:27 AM
P: 3
1. The problem statement, all variables and given/known data
Integrating 1/xlnx by parts...
2. Relevant equations
Find the integral of 1/xlnx

The question asks to solve by substitution, which I can do and results in ln(ln(x)) + c

It then asks to compute using integration by parts, and then to explain how it can be true (because it will compute something different to substitution).

3. The attempt at a solution
I = uv - int (v dU)

let u= 1/lnx du = 1/x(lnx)^2
let dv = 1/x, v = lnx

Sub into the parts formula

I = lnx* 1/lnx - int (lnx/x(lnx)^2)
I = lnx/lnx - int (1/xlnx) <--- what we started with
I = 1 - int (1/xlnx) This is 1 - I, the integral we began with...

I've bene shown this trick where you can go..
I = 1- I
2I = 1
I = 1/2

I'm not sure if this is correct, but I would appreciate any help

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



2. Relevant equations



3. The attempt at a solution
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n.karthick
#2
Sep13-10, 08:35 AM
P: 241
It is correct
ductruffe
#3
Sep13-10, 08:37 AM
P: 3
Thank you for the quick reply. I'm also asked to explain how I = 1/2 can be true, when using substitution yields ln(ln(x)). This is basically where I am stuck.

Thank you

n.karthick
#4
Sep13-10, 09:01 AM
P: 241
Integrating 1/xln(x) using integration by parts

ln(ln(x)) is correct.
Regarding Integration by parts, you have missed -ve sign in differentiating u
It will be
I=1+I
resulting in 1=0 which is wrong.
jgens
#5
Sep13-10, 11:04 AM
P: 1,622
Quote Quote by n.karthick View Post
ln(ln(x)) is correct.
Regarding Integration by parts, you have missed -ve sign in differentiating u
It will be
I=1+I
resulting in 1=0 which is wrong.
No, it results in 1 = C.
ductruffe
#6
Sep13-10, 04:29 PM
P: 3
I can see I've missed the negative which changes it quite a bit.

Is the integration by parts correct for C=1? It hasn't really solved the integral, or am I missing something here?
jgens
#7
Sep13-10, 04:37 PM
P: 1,622
Typically, you'll see this problem as an example of why it's so important to remember the constant of integration, because otherwise you end up with nonsense like 1 = 0.


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