integral ln(1-x)dx , tolong yang bisa bantu

jawab

u = 1- x
u' = -  dx

∫ ln ( 1- x)  dx
=  - ∫  ln (u) du
= - { u ln u - u )
= - u ln u + u
= -(1-x) ln (1- x) + (1- x)
= - ln (1-x) + x ln (1-x) + (1 - x) + c