∫[-1/2,0]ln(1-x)dx 怎么算到结果啊

∫(-1/2-->0) ln(1 - x) dx
= xln(1 - x) - ∫(-1/2-->0) x d[ln(1 - x)] 0) x · 1/(1 - x) · (- 1) dx
= (1/2)ln(3/2) - ∫(-1/2-->0) [(- x + 1) - 1]/(1 - x) dx
= (1/2)ln(3/2) - [x + ln(1 - x)]
= (1/2)ln(3/2) - [(0 + ln(1 - 0)) - (- 1/2 + ln(1 + 1/2))]
= (1/2)ln(3/2) + ln(3/2) - 1/2
= (3/2)ln(3/2) - 1/2 ≈ 0.1082