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求定积分dx/(e^x+1+e^3-x) 上限正无穷,下限0
人气:279 ℃ 时间:2020-03-28 15:42:23
解答
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)-1/4怎么来的???(e^x+1/2)^2=e^2x+e^x+1/4诶,我把题目打错误了,e^(x+1)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
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