Question

how do you take the integral of (lnx)/(x^2)?

Answer 1

INTlnx* x^-2
try integration by parts

Answer 2

What should my u and v be?

Answer 3

try u = lnx so du = 1/x
and dv = x^-2 so v = - x ^-1 or -1/x

Answer 4

lets put lnx as u
so it'll be int ( u/x du)

Answer 5

INT = (-1/x)* lnx - INT(-1/x * 1/x dx
= (-1/x) ln x + INT x^-2 dx
= -1/x) ln x - 1/x
= -1/x (ln x+ 1) + c