admin百科知識 2022-02-13 23:30:50 ∫[1→+∞] 1(e^x+e^(2-x))dx=________________.∫[1→ ∞] 1/(e^x e^(2-x))dx=________________.網友廻答：匿名網友答[dá]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,則有 ∫[1→ ∞] 1/(e^x e^2)de^x==∫[e→ ∞] 1/(t^2 e^2)dt=1/e∫[e→ ∞]... db標簽 生活常識_百科知識_各類知識大全»∫[1→+∞] 1(e^x+e^(2-x))dx=________________.
![∫[1→+∞] 1(e^x+e^(2-x))dx=________________.,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw== "∫[1→+∞] 1(e^x+e^(2-x))dx=________________.,第1張")
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![∫[1→+∞] 1(e^x+e^(2-x))dx=________________.,第2張](/statics/default/images/niming.jpg "∫[1→+∞] 1(e^x+e^(2-x))dx=________________.,第2張")
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