求{1/[e^x+e^(-x)]}dx,0

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求{1/[e^x+e^(-x)]}dx,0求{1/[e^x+e^(-x)]}dx,0求{1/[e^x+e^(-x)]}dx,0积分号我就不打了,其实我是不会打~{1/[e^x+e^(-x)]}dx=e

求{1/[e^x+e^(-x)]}dx,0
求{1/[e^x+e^(-x)]}dx,0

求{1/[e^x+e^(-x)]}dx,0
积分号我就不打了,其实我是不会打~
{1/[e^x+e^(-x)]}dx
=e^x/[(e^x)²+1]dx
=1/[(e^x)²+1]d(e^x)
=arctan(e^x)
所求积分
=arctan(e^1)-arctan(e^0)
=arctane-arctan1
=arctane-(派/4)

{1/[e^x+e^(-x)]}dx
上下同乘以e^x
={e^x/[e^(2x)+1]}dx
因为:e^xdx=d(e^x)
={1/[(e^x)^2+1]}d(e^x)
由:[1/(1+x^2)]dx=arctanx
=arctan(e^x)
因为:0=arctan(e^x)|(0到1)
=arctan(e)-arctan1