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Integrating 1/xln(x) using integration by partsby ductruffe
Tags: integration parts ln 
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#1
Sep1310, 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 


#2
Sep1310, 08:35 AM

P: 241

It is correct



#3
Sep1310, 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 


#4
Sep1310, 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. 


#5
Sep1310, 11:04 AM

P: 1,622




#6
Sep1310, 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? 


#7
Sep1310, 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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