# Integration by parts

1. May 10, 2007

### Rasine

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?

2. May 10, 2007

### cristo

Staff Emeritus
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

Last edited: May 10, 2007
3. May 10, 2007

### neutrino

How about integration-by-parts once again? :)

4. May 10, 2007

### quasar987

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. May 10, 2007

### Rasine

ohhh yes......do integration by part again........

thank you!