积分 dx/[e^x+e^(2-x)]

令t=e^x,则dt=e^x*dx=tdx
dx/[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)
=arctan(u)/e+C1
=arctan(t/e)/e+C2
=arctan(e^x/e)/e+C3=arctan[e^(x-1)]/e+C