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Integration by parts 
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#1
May1007, 02:50 PM

P: 214

intergrate (ln(x))^2
so i set u=(lnx)^2...which makes du=2lnx(1/x) then i set dv=dx...which makes v=x according to the formula for integration by parts i have x(lnx)^2 integral x(2lnx)(1/x) simplifying it i get x(ln)^22intergral lnx and here is where i am stuck....what i the integral of lnx? 


#2
May1007, 02:53 PM

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P: 8,317

The derivative of (lnx)^2=(2lnx)/x.
[edit: of course that's you've written... i glanced and say ln(1/x)... sorry ] A hint for integrating lnx; use parts, taking dv=dx and u=lnx 


#3
May1007, 02:54 PM

P: 2,046

How about integrationbyparts once again? :)



#4
May1007, 03:00 PM

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P: 4,771

Integration by parts
xln(x)  x looks good from where I'm standing.
I just wondered "what function gives ln(x) when differentiated? Well ln(x)' = 1/x. So what if I try xln(x)? Now I get ln(x) + 1. So I need to add something to the mix that gives 1 when differentiated." Hence xln(x)  x. 


#5
May1007, 03:06 PM

P: 214

ohhh yes......do integration by part again........
thank you! 


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