∫[-1/2,0]ln(1-x)dx 怎么算到结果啊

网友回答

∫(-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
======以下答案可供参考======
供参考答案1:
设t=1-x 当x=-1/2 t=3/2 , x=0 t=1
因此 ∫[-1/2,0]ln(1-x)dx
=∫[3/2,1]lntdt
=-∫[1,3/2]lntdt
=-[ln(3/2)-ln1]
=ln2-ln3
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