题目

积分 dx/[e^x+e^(2-x)]
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提问时间：2021-02-25

答案

令t=e^x,则dt=e^x*dx=tdxdx/[e^x+e^(2-x)]=dx/[t+(e^2/t)]=tdx/(t^2+e^2)=dt/(t^2+e^2)令t/e=u,t=eu,则dt=edu,dt/(t^2+e^2)=edu/[e^2(1+u^2)]=du/e(1+u^2)∫dx/[e^x+e^(2-x)]=∫du/e(1+u^2)=(1/e)∫du/(1+u^2)=arcta...