积分 dx/[e^x+e^(2-x)]请看清楚题目作答

dx/[e^x+e^(2-x)]
=e^x-e^(2-x)

∫[e^x+e^(2-x)]dx
=∫e^xdx+∫e^(2-x)dx
=e^x+C1-∫e^(2-x)d(2-x)
=e^x+C1-e^(2-x)+C2
=e^x-e^(2-x)+c

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...