∫e^x(x-1)/x²dx
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解:∫(-π/4到π/4)
(cosx)2/[1+e^(-x)]dx
=∫(-π/4到0)
(cosx)2/[1+e^(-x)]dx+∫(0到π/4)
(cosx)2/[1+e^(-x)]dx
对第一个积分式,令t=-x代换下,有:
∫(-π/4到0)
(cosx)2/[1+e^(-x)]dx
(
t=-x,则dx=-dt)
=∫(π/4到0)
(cost)2/[1+e^t](-dt)
=∫(0到π/4)
(cost)2/[1+e^t]dt
=∫(0到π/4)
(cosx)2/[1+e^x]dx
故:原积分式
=∫(0到π/4)
(cosx)2/[1+e^x]dx+∫(0到π/4)
(cosx)2/[1+e^(-x)]dx
=∫(0到π/4)
[(cosx)2/(1+e^x)+(cosx2)e^x/(e^x+1)]dx
=∫(0到π/4)
(1+cos2x)/2dx
=(x+1/2*sin2x)|(0到π/4)
=π/4+1/2
(cosx)2/[1+e^(-x)]dx
=∫(-π/4到0)
(cosx)2/[1+e^(-x)]dx+∫(0到π/4)
(cosx)2/[1+e^(-x)]dx
对第一个积分式,令t=-x代换下,有:
∫(-π/4到0)
(cosx)2/[1+e^(-x)]dx
(
t=-x,则dx=-dt)
=∫(π/4到0)
(cost)2/[1+e^t](-dt)
=∫(0到π/4)
(cost)2/[1+e^t]dt
=∫(0到π/4)
(cosx)2/[1+e^x]dx
故:原积分式
=∫(0到π/4)
(cosx)2/[1+e^x]dx+∫(0到π/4)
(cosx)2/[1+e^(-x)]dx
=∫(0到π/4)
[(cosx)2/(1+e^x)+(cosx2)e^x/(e^x+1)]dx
=∫(0到π/4)
(1+cos2x)/2dx
=(x+1/2*sin2x)|(0到π/4)
=π/4+1/2
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