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积分号xlnx/(1+x^2)^2
人气:326 ℃ 时间:2020-06-05 22:48:37
解答
分部积分啦!∫xlnx/[(1+x^2)^2]dx=(-1/2)∫lnxd(1/(1+x^2))=(-1/2)lnx/(1+x^2)+(1/2)∫1/[(1+x^2)*x]dx=(-1/2)lnx/(1+x^2)+(1/2)∫x/[(1+x^2)*x^2]dx=(-1/2)lnx/(1+x^2)+(1/4)∫1/[(1+x^2)*x^2]d(x^2)=(-1/2)lnx/(1+...
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