作业帮数学题第六题第8题
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数学题第六题第8题
数学题第六题第8题
数学题第六题第8题
⑥令t=1-x dx=-dt
∫(0,1) x/(e^x+e^(1-x))dx
=∫(0,1) (1-t)/(e^(1-t)+e^t)dt
=∫(0,1) dt/(e^t+e^(1-t)) - ∫(0,1) t/(e^(1-t)+e^t)dt
所以原式=(1/2)*∫(0,1) 1/(e^x+e^(1-x))dx
=(1/2)*∫(0,1) e^x/(e^2x+e)dx
=(1/2)*∫(0,1) d(e^x)/(e^2x+e)
=(1/2)*(1/√e)*arctan(e^x/√e)|(0,1)
=(1/2√e)*[arctan(√e)-arctan(1/√e)]
=(1/2√e)*[2arctan(√e)-π/2]
=(1/√e)*arctan(√e)-π/4√e
⑧原式=(1/e)*∫(1,+∞) e^x/(e^2x+e^2)dx
=(1/e)*∫(1,+∞) d(e^x)/(e^2x+e^2)
=(1/e^2)*arctan[e^(x-1)]|(1,+∞)
=π/2e^2-π/4e^2
=π/4e^2