Question

∫[1→+∞] 1/(e^x+e^(2-x))dx=________________.

Answer 1

答案：π/4e；

∫[1→+∞] 1/(e^x+e^(2-x))dx=∫[1→+∞] e^x/(e^2x+e^2)dx=∫[1→+∞] 1/(e^2x+e^2)de^x

不妨令t=e^x,则有

∫[1→+∞] 1/(e^x+e^2)de^x==∫[e→+∞] 1/(t^2+e^2)dt=1/e∫[e→+∞] 1/[(t/e)^2+1]d(t/e)

==1/e*arctan(t/e)[e→+∞]=1/e(π/2-π/4)=π/4e