求下列不定积分 ∫(arctan e^x)/(e^2x)dx
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亲
原式= ∫(arctan e^x)/(e^x)d(1/e^x)
令1/e^x=t = ∫(arctan 1/t)/t dt
= (-1/2) ∫(arctan 1/t) d(t^2)
= (-1/2) [(t^2)(arctan 1/t) - ∫(t^2)d(arctan 1/t)]
= (-1/2) [(t^2)(arctan 1/t) + ∫(t^2)/(1+t^2)dt]
= (-1/2) [(t^2)(arctan 1/t) + ∫(t^2+1-1)/(1+t^2)dt
= (-1/2) {(t^2)(arctan 1/t) +∫1-[1/(1+t^2)]dt}
= (-1/2)[e^-(2x)*arctane^x-arctane^(-x)+e^(-x)]+C
咨询记录 · 回答于2021-11-09
求下列不定积分 ∫(arctan e^x)/(e^2x)dx
亲~您好，很高兴能为您服务，您的问题我看到啦，正在为您整理资料，请耐心等待
亲原式= ∫(arctan e^x)/(e^x)d(1/e^x)令1/e^x=t = ∫(arctan 1/t)/t dt= (-1/2) ∫(arctan 1/t) d(t^2)= (-1/2) [(t^2)(arctan 1/t) - ∫(t^2)d(arctan 1/t)]= (-1/2) [(t^2)(arctan 1/t) + ∫(t^2)/(1+t^2)dt]= (-1/2) [(t^2)(arctan 1/t) + ∫(t^2+1-1)/(1+t^2)dt= (-1/2) {(t^2)(arctan 1/t) +∫1-[1/(1+t^2)]dt}= (-1/2)[e^-(2x)*arctane^x-arctane^(-x)+e^(-x)]+C
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