Integral of (ln^2)X

galadriel100

Found this question in a calculus book from 1977 and I don't know how to do it..

Find the integral of (ln^2)x

Reply 1

jpowell

You just need to use the chain rule, to get 2*log(x)/x.

Reply 2

meef cheese

jpowell

You just need to use the chain rule, to get 2*log(x)/x.

Integral, not derivative.

Reply 3

meef cheese

Use parts. Let u = lnx and dv/dx = lnx and remember that

\int ln x\; dx = x (ln x -1)

and it should be easy.

Reply 4

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

Reply 5

jpowell

e-unit

Integral, not derivative.

Doh,

Reply 6

meef cheese

driving_seat

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