The Student Room Group
Reply 1
You just need to use the chain rule, to get 2*log(x)/x.
jpowell
You just need to use the chain rule, to get 2*log(x)/x.

Integral, not derivative.
Use parts. Let u = lnx and dv/dx = lnx and remember that lnx  dx=x(lnx1) \int ln x\; dx = x (ln x -1) and it should be easy.
well if u mean ln²x then u'll have to do by applying by parts

I=∫ln²x dx

u=ln²x v'=∫1 dx
u'=(2lnx)/x v=x

I=(ln²x)(x)-∫(2lnx)/x*x dx
I=xln²x-2∫lnx dx

Applying By Parts again

u=lnx v'=∫1 dx
u'=1/x v=x

I=xln²x-2[(x)(lnx) -∫1 dx]
I=xln²x-2xlnx -2x +c
Reply 5
e-unit
Integral, not derivative.


Doh, :p:
driving_seat
well if u mean ln²x then u'll have to do by applying by parts

I=∫ln²x dx

u=ln²x v'=∫1 dx
u'=(2lnx)/x v=x

I=(ln²x)(x)-∫(2lnx)/x*x dx
I=xln²x-2∫lnx dx

Applying By Parts again

u=lnx v'=∫1 dx
u'=1/x v=x

I=xln²x-2[(x)(lnx) -∫1 dx]
I=xln²x-2xlnx -2x +c

I'm sure it gives your life meaning to realise that you can do the question, but it's not really helpful to the OP to post a full solution. And your answer is wrong. It should be I = xln²x - 2xlnx + 2x + C
like DUHHH....