2019-01-05 20:53发布
设y=e^x，则x=lny。
∫dx/(e^x+1)
=∫dlny/(y+1)
=∫dy/[y(y+1)]
=∫[1/y-1/(y+1)]dy
=lny-ln(y+1)+c
=x-ln(e^x+1)+c
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设y=e^x，则x=lny。
∫dx/(e^x+1)
=∫dlny/(y+1)
=∫dy/[y(y+1)]
=∫[1/y-1/(y+1)]dy
=lny-ln(y+1)+c
=x-ln(e^x+1)+c
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