Question

广义积分∫ln(1-x^2)dx(0到1)

Answer 1

∫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+x)]dx
=xln(1-x^2)-2x+ln(1+x)-ln(1-x)+C
积分限为0≤x≤1，则
∫<0,1>ln(1-x^2)dx
=[1*ln(1-1^2)-2*1+ln(1+1)-ln(1-1)]-[0*ln(1-0^2)-2*0+ln(1+0)-ln(1-0)]
=[ln0-2+ln2-ln0]-[0-0+0-0]
=ln2-2

Answer 2

=∫ln(1-x)dx+∫ln(1+x)d(1+x)
=-∫ln(1-x)d(1-x)+[(1+x)ln(1+x)-x](0,1) =[-(1-x)ln(1-x)-x](0,1)+ln2-1
=-1-lim(x趋于1)(1-x)ln(1-x)+ln2-1 =ln2-2

Answer 3

直接用分部积分