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求xe^xdx\(e^x-1)^2的不定积分
人气:226 ℃ 时间:2020-01-29 03:19:44
解答
令a=e^xx=lnadx=da/a原式=∫alna*(da/a)/(a-1)^2=∫lnada/(a-1)^2=∫lnad[-1/(a-1)]=lna[-1/(a-1)]-∫[-1/(a-1)]dlna=-lna/(a-1)+∫[1/a(a-1)]da=-lna/(a-1)+∫[da/a-da/(a-1)]=-lna/(a-1)+lna-ln|a-1|+C=-x/(e^x-1)...
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