How to find integral of ln(lnx) + 1/lnx

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

So, ∫ (ln(ln x) + 1/ln x) dx
= ∫ ln(ln x) dx + ∫ dx/ln x
= [x ln(ln x) - ∫ x * dx/(x ln x)] + ∫ dx/ln x, via integration by parts
= [x ln(ln x) - ∫ dx/ln x] + ∫ dx/ln x
= x ln(ln x) + C.

I hope this helps!

The integral of ln|x| is found by multiplying it by x and then tinkering with it until it gives the right derivative.
Try multiplying ln(ln(x)) by x and keeping x > 1. Or better multiply ln| ln|x| | by x with x not = 0.