Find the integral of [ ln ( x ) ] / [ x ^ 2 ]?

Answer 1

use integration by parts:

let u=ln(x), so du=1/x dx. Let dv=x^-2, so v=-x^-1. Hence you have the integral = uv-integral(v*du) = -ln(x)/x+integral(x^-2 dx) =-ln(x)/x -1/x +c

Answer 2

Integrate by parts
= -1/x *ln x +int 1/x^2 dx = -1/x(ln x+1)+C