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求{1/[e^x+e^(-x)]}dx,0
人气:364 ℃ 时间:2020-05-06 01:10:54
解答
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)
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