3jkc0
幼苗
共回答了19个问题采纳率：84.2% 举报
凑微+分部积分+变量替换
记I=∫ (1~+∞)arctanx/(x^2) dx
=-∫ (1~+∞)arctanxd(1/x )
=-(1/x)arctanx|(1,+∞)+∫ (1~+∞)1/[x(1+x^2)]dx
=π/4+∫ (1~+∞)1/[x(1+x^2)]dx
令1/x=t.则∫ (1,+∞)1/[x(1+x^2)]dx=∫(0~1)t/(1+t^2)dt=(1/2)ln(1+t^2)|(0,1)=(1/2)ln2
所以I=π/4+(1/2)ln2
1年前
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