Question

求{1/[e^x+e^(-x)]}dx,0<x<1的定积分

Answer 1

解：利用积分公式：积分（f(x)+g(x)dx=积分f(x)dx+积分g(x)dx
令f(x)=x,g(x)=e^x
积分（x+e^x)dx=积分xdx+积分e^xdx
=1/2x^2/(0
1)+e^x/(0
1)
=1/2(1^2-0^2)+e^1-e^0
=1/2(1-0)+e-1
=1/2+e-1
=e-1/2
答：答案是e-1/2.

Answer 2

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)