题目

求{1/[e^x+e^(-x)]}dx,0

提问时间：2020-11-03

答案

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)