Question

求定积分dx/(e^x+1+e^3-x) 上限正无穷，下限0

Answer 1

定积分dx/(e^x+1+e^3-x) 上限正无穷，下限0
=∫(0,+∞)e^x/(e^2x+e^x+e^3)dx
=∫(0,+∞)e^x/((e^x+1/2)^2+e^3-1/4)dx
=1/√(e^3-1/4)*arctan(e^x+1/2)/√(e^3-1/4)|(0,+∞)
=π/2√(e^3-1/4)-1/√(e^3-1/4)*arctan(3/2)/√(e^3-1/4)

追问

-1/4怎么来的？？？

追答

(e^x+1/2)^2=e^2x+e^x+1/4

追问

诶，我把题目打错误了，e^(x+1)

追答

定积分dx/(e^x+1+e^3-x) 上限正无穷，下限0
=∫(0,+∞)e^x/(e*e^2x+e^3)dx
=1/e*∫(0,+∞)e^x/(e^2x+e^2)dx
=1/e*∫(0,+∞)de^x/(e^2x+e^2)
=1/e*1/e*arctane^x/e|(0,+∞)
=π/2e^2-1/e^2*arctan1/e