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∫(上限ln2,下限0) e^x/1+e^2x dx
人气:107 ℃ 时间:2020-03-27 10:53:03
解答
∫[0,ln2] e^x/(1+e^2x )dx
=∫[0,ln2] 1/(1+e^2x )de^x
=arctane^x[0,ln2]
=arctan2-π/4
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