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∫上限正无穷下限0 arctanx/(1+x^2)^(3/2)dx
人气:154 ℃ 时间:2020-05-21 08:22:41
解答
∫(0->+∞) arctanx / (1+x^2)^(3/2)dxletx= tanadx= (seca)^2 dax=0,a=0x=+∞,a=π/2∫(0->+∞) arctanx / (1+x^2)^(3/2)dx=∫(0->π/2) [a / (seca)^3 ] (seca)^2 da=∫(0->π/2) acosa da=∫(0->π/2) adsina=[as...
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