∫(√1+e^x)dx
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∫(√1+e^x)dx
∫(√1+e^x)dx
∫(√1+e^x)dx
令 √(1+e^x) = u, 则 e^x=u^2-1, x=ln(u^2-1), dx= 2udu/(u^2-1)
I = ∫ √(1+e^x)dx = ∫ 2u^2du/(u^2-1) = 2 ∫ [1+1/(u^2-1)]du
= 2u + ∫ [1/(u-1)-1/(u+1)]du = 2u + ln |(u-1)/(u+1)| + C
= 2√(1+e^x) + ln |[√(1+e^x)-1]/[√(1+e^x)+1)]| + C
= 2√(1+e^x) + 2ln[√(1+e^x)-1] - x + C
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