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Integration by parts

by Rasine
Tags: integration, parts
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Rasine
#1
May10-07, 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)^2-2intergral lnx


and here is where i am stuck....what i the integral of lnx?
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cristo
#2
May10-07, 02:53 PM
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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
neutrino
#3
May10-07, 02:54 PM
P: 2,046
How about integration-by-parts once again? :)

quasar987
#4
May10-07, 03:00 PM
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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.
Rasine
#5
May10-07, 03:06 PM
P: 214
ohhh yes......do integration by part again........

thank you!


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