求(1-lnx)/(x-lnx)^2的定积分

Question

如题所述

Answer 1

(x-lnx) ' = 1 - 1/x,
∫ [(1- lnx) / (x-lnx)^2 ] dx = ∫ [(x- lnx) - x * (1 - 1/x) ] /(x-lnx)^2 ] dx
= ∫ (-x) * (1 - 1/x) / (x-lnx)^2 ] dx + ∫ 1/(x- lnx) dx
= ∫ x d [1/(x-lnx)] + ∫ 1/(x- lnx) dx
= x / (x-lnx) - ∫ 1/(x- lnx) dx + ∫ 1/(x- lnx) dx
= x / (x-lnx) + C