(ln(x))^2 dx, integrate.

Answer 1

use integration by parts. That is just the product rule in reverse.

d(uv) = u dv + v du
u dv = d(uv) - v du
∫u dv = ∫d(uv) - ∫v du
∫u dv = uv - ∫v du

So, here we just let

u = (lnx)^2
dv = dx

du = 2lnx * 1/x dx = (2lnx)/x
v = x

∫(lnx)^2 dx = x(lnx)^2 - ∫2lnx dx

Now repeat, this time letting

u = lnx
dv = 2dx