How to integrate (In x)^2 ?

Question

How to integrate (In x)^2 ? thanks

Answer 1

Use integration by parts (Sorry I haven't access to typing the correct notation: I'll use (Itl) to mean the integral sign.)

(Itl) v du = uv - (Itl) u dv
In this case, as v = (ln x)^2, we have dv = 2 (ln x).(1/x) dx

Therefore (Itl) (ln x)^2 dx
= x (ln x)^2 - (Itl) x. 2(ln x).(1/x) dx
= x (ln x)^2 - (Itl) 2 (ln x) dx
Apply integration by parts again to the second term, getting
x (ln x)^2 - 2x ln x + (itl) 2x.(1/x) dx
= x (ln x)^2 - 2x ln x + (Itl) 2 dx
= x (ln x)^2 - 2x ln x + 2x,

which I guess is the same as the answer someone else found in a book of integrals.

Answer 2

Integral of (ln x)^2 dx = x[2 - 2lnx + (ln x)^2] + C

Ashish- Chain rule is for differentiation not integration

Answer 3

integral(ln(x)^2) = x(ln(x)^2) - 2xln(x) + 2x + c, using the method of Integration by Parts.

Basically, intg(udv) = uv - intg(vdu)

have u = (ln(x)^2), du = 2ln(x)*(1/x)
dv = dx, v = x

so you get: intg((ln(x)^2)dx) = x*(ln(x)^2) - intg(2ln(x))

Then you need to reapply Integration by Parts for the last integral:

u = ln(x), du = (1/x)
dv = dx, v = x

intg(2ln(x)) = xln(x) - intg(1) = xln(x) - x + c

plug this back into your equation to get the result I listed above.


...and technically, Integration by Parts IS a modification of the Product Rule. Just switch a couple terms to different sides of the equation and put integral signs in front of every term.

Answer 4

Use integration by parts. Twice.

Answer 5

I agree to the solution given by Ashish P.

However, I may add as a remark that, every function of 'x' is not necessarily integrable!

Answer 6

write it as 1.(ln x)^2
and then use the product rule!