求这道题的详细过程
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说明:^2——表示平方
∫xln[(1+x)/(1-x)]dx
=∫xln(1+x)dx-∫xln(1-x)dx
=1/2∫ln(1+x)dx^2-1/2∫ln(1-x)dx^2
=1/2x^2ln(1+x)-1/2∫x^2/(1+x)dx-1/2x^2ln(1-x)+1/2∫-x^2/(1-x)dx
=1/2x^2[ln(1+x)-ln(1-x)]-1/2∫[(x^2-1)+1]/(1+x)dx+1/2∫[(1-x^2)-1]/(1-x)dx
=1/2x^2ln[(1+x)/(1-x)]-1/2∫(x-1)dx-1/2∫1/(1+x)dx+1/2∫(1+x)dx+1/2∫1/(1-x)d(-x)
=1/2x^2ln[(1+x)/(1-x)]-1/4(x-1)^2-1/2ln(1+x)+1/4(1+x)^2+1/2ln(1-x)
=1/2x^2ln[(1+x)/(1-x)]+1/4[(1+x)^2-(1-x)^2]+1/2[ln(1-x)-ln(1+x)]
=1/2x^2ln[(1+x)/(1-x)]+1/4[(1+x)+(1-x)][(1+x)-(1-x)]+1/2ln[(1-x)/(1+x)]
=1/2x^2ln[(1+x)/(1-x)]+x-1/2ln[(1+x)/(1-x)]
=1/2ln[(1+x)/(1-x)](x^2-1)
∫xln[(1+x)/(1-x)]dx
=∫xln(1+x)dx-∫xln(1-x)dx
=1/2∫ln(1+x)dx^2-1/2∫ln(1-x)dx^2
=1/2x^2ln(1+x)-1/2∫x^2/(1+x)dx-1/2x^2ln(1-x)+1/2∫-x^2/(1-x)dx
=1/2x^2[ln(1+x)-ln(1-x)]-1/2∫[(x^2-1)+1]/(1+x)dx+1/2∫[(1-x^2)-1]/(1-x)dx
=1/2x^2ln[(1+x)/(1-x)]-1/2∫(x-1)dx-1/2∫1/(1+x)dx+1/2∫(1+x)dx+1/2∫1/(1-x)d(-x)
=1/2x^2ln[(1+x)/(1-x)]-1/4(x-1)^2-1/2ln(1+x)+1/4(1+x)^2+1/2ln(1-x)
=1/2x^2ln[(1+x)/(1-x)]+1/4[(1+x)^2-(1-x)^2]+1/2[ln(1-x)-ln(1+x)]
=1/2x^2ln[(1+x)/(1-x)]+1/4[(1+x)+(1-x)][(1+x)-(1-x)]+1/2ln[(1-x)/(1+x)]
=1/2x^2ln[(1+x)/(1-x)]+x-1/2ln[(1+x)/(1-x)]
=1/2ln[(1+x)/(1-x)](x^2-1)
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