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求定积分,上限为1,下限为0,求不定积分[x平方/(1+x平方)]dx
人气:276 ℃ 时间:2020-01-29 16:26:23
解答
原式=∫[0,1](1+x^2-1)dx/(1+x^2)
=∫[0,1]dx-∫[0,1]dx/(1+x^2)
=x[0,1]-arctanx[0,1]
=1-0-(π/4-0)
=1-π/4.
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