I = ∫e^x(e^x-1)dx/{e^x[e^(2x)+4)]}
= ∫(e^x-1)de^x/{e^x[e^(2x)+4)]} 令 u = e^x
= ∫(u-1)du/[u(u^2+4)] = (1/4)∫[(u+4)/(u^2+4)-1/u]du
= (1/4)∫[u/(u^2+4)+4/(u^2+4)-1/u]du
= (1/4)[(1/2)ln(u^2+4) + 2arctan(u/2) - lnu] + C
= (1/8)ln[e^(2x)+4)] + (1/2)arctan(e^x/2) - x/4 + C
= ∫(e^x-1)de^x/{e^x[e^(2x)+4)]} 令 u = e^x
= ∫(u-1)du/[u(u^2+4)] = (1/4)∫[(u+4)/(u^2+4)-1/u]du
= (1/4)∫[u/(u^2+4)+4/(u^2+4)-1/u]du
= (1/4)[(1/2)ln(u^2+4) + 2arctan(u/2) - lnu] + C
= (1/8)ln[e^(2x)+4)] + (1/2)arctan(e^x/2) - x/4 + C
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