广义积分∫ln(1-x^2)dx(0到1)
人气：223 ℃ 时间：2020-04-29 09:46:03
解答
∫ln(1-x^2)dx=xln(1-x^2)-∫xdln(1-x^2)=xln(1-x^2)-∫x/(1-x^2)*(-2x)dx=xln(1-x^2)-2∫(-x^2)/(1-x^2)dx=xln(1-x^2)-2∫(1-x^2-1)/(1-x^2)dx=xln(1-x^2)-2∫dx+2∫1/(1-x^2)dx=xln(1-x^2)-2∫dx+∫[1/(1-x)+1/(1+...
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