∫dx/(1+e^x)怎么算?
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∫dx/(1+e^x)怎么算?
![∫dx/(1+e^x)怎么算?](/uploads/image/z/15262443-27-3.jpg?t=%E2%88%ABdx%2F%281%2Be%5Ex%29%E6%80%8E%E4%B9%88%E7%AE%97%3F)
令a=1/(1+e^x)
e^x=1/a-1=(1-a)/a
x=ln[(1-a)/a]
dx=[a/(1-a)]*[-a-(1-a)]/a^2 da=-1/(a-a^2) da
所以原式=∫a*[-1/(a-a^2)]da
=∫1/(a-1)da
=∫1/(a-1)d(a-1)
=ln|a-1|+C
=ln|1/(1+e^x)-1|+C
=ln[e^x/(1+e^x)]+C
e^x=1/a-1=(1-a)/a
x=ln[(1-a)/a]
dx=[a/(1-a)]*[-a-(1-a)]/a^2 da=-1/(a-a^2) da
所以原式=∫a*[-1/(a-a^2)]da
=∫1/(a-1)da
=∫1/(a-1)d(a-1)
=ln|a-1|+C
=ln|1/(1+e^x)-1|+C
=ln[e^x/(1+e^x)]+C