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(x-1)/(x^2+3)的不定积分、、
人气:294 ℃ 时间:2020-02-05 17:07:46
解答
∫[(x-1)/(x^2+3)]dx
=∫[x/(x^2+3)]dx-∫[1/(x^2+3)]dx
=(1/2)∫[1/(x^2+3)]d(x^2+3)-(1/√3)∫{1/[(x/√3)^2+1]}d(x/√3)
=(1/2)ln(x^2+3)-(1/√3)arctan(x/√3)+C.
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