∫[1→+∞] 1/(e^x+e^(2-x))dx=________________.
问题描述:
∫[1→+∞] 1/(e^x+e^(2-x))dx=________________.
答
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→+∞]...