求{1/[e^x+e^(-x)]}dx,0
问题描述：
求{1/[e^x+e^(-x)]}dx,0
答
{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)-arctan1
答
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)