Question

求不定积分∫xex/√ex-2dx

Answer 1

let
e^(x/2) = √2secu
(1/2)e^(x/2) dx =√2secu.tanu du
dx = 2√2 tanu du
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∫xe^x/√(e^x-2) dx
=2∫xd√(e^x-2)
=2x.√(e^x-2) - 2∫√(e^x-2) dx
=2x.√(e^x-2) - 2∫ √2tanu . ( 2√2 tanu du )
=2x.√(e^x-2) - 8∫ (tanu)^2 du
=2x.√(e^x-2) - 8∫ [(secu)^2-1] du
=2x.√(e^x-2) - 8(tanu -u) +C
=2x.√(e^x-2) - 8{ √(e^x-2) /√2 -arctan[√(e^x-2) /√2] } +C

Answer 2

∫xe^x/(e^x +1)^2dx
= -∫xd[1/(1+e^x)]
= -x/(1+e^x)+∫[1/(1+e^x)]dx
= -x/(1+e^x)+∫[(1+e^x-e^x)/(1+e^x)]dx
= -x/(1+e^x)+∫1dx-∫(1/(1+e^x))d(1+e^x)
=-x/(1+e^x)+x-ln(1+e^x)+C