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求定积分 ∫(上1下0)1/(x^2-2x-3)dx
人气:107 ℃ 时间:2020-04-03 02:02:42
解答
∫[0,1] 1/(x^2-2x-3)dx
=1/5∫[0,1] [1/(x-3)-1/(x+2)]dx
=1/5[ln|x-3|-ln|x+2|] [0,1]
=1/5(ln2-ln3-ln2+ln2)
=0
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