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

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

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

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