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∫lny-ln（1-y）dy积分区间是1/2至1
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∫lny-ln（1-y）dy 积分区间是1/2至1
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答：
先用分部积分法求不定积分
∫ lny-ln(1-y) dy
=y*[lny-ln(1-y)] -∫ y* [1/y+1/(1-y)] dy
=ylny-yln(1-y) - ∫ 1/(1-y) dy
=yln | y |-yln |1-y| +ln |y-1|+C
=yln |y|+(1-y) ln|1-y|+C
定积分=(1/2→1) [ yln | y | +(1- y)ln | y-1 | ]
=(0+0) - [ (1/2)ln(1/2)+(1/2)ln(1/2) ]
=-ln(1/2)
=ln2
先用分部积分法求不定积分
∫ lny-ln(1-y) dy
=y*[lny-ln(1-y)] -∫ y* [1/y+1/(1-y)] dy
=ylny-yln(1-y) - ∫ 1/(1-y) dy
=yln | y |-yln |1-y| +ln |y-1|+C
=yln |y|+(1-y) ln|1-y|+C
定积分=(1/2→1) [ yln | y | +(1- y)ln | y-1 | ]
=(0+0) - [ (1/2)ln(1/2)+(1/2)ln(1/2) ]
=-ln(1/2)
=ln2
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