Verified answer

∫log(y) dy/y

integrate by parts

let u = log(y): dv = dy/y

du = dy/y : v = log(y)

∫u dv = uv - ∫v du

∫log(y) dy/y = [log(y)]^2 - ∫log(y) dy/y

add ∫log(y) dy/y both sides

2 ∫log(y) dy/y = [log(y)]^2

divide by 2

∫log(y) dy/y = (1/2) [log(y)]^2 + c

½ln²(y) + C

Basic problem utilizing the Integration by Parts technique