Question

瑕积分求解

Answer 1

1/(1-x^2) = 1/[(1-x)(1+x)]
ln[1/(1-x^2)] = -ln(1-x) -ln(1+x)
∫ln[1/(1-x^2)] dx
=-∫[ ln(1-x) + ln(1+x) ]dx
=-x[ ln(1-x) + ln(1+x) ] + ∫x[ -1/(1-x) +1/(1+x) ]dx
=-x[ ln(1-x) + ln(1+x) ] + ∫[ -x/(1-x) +x/(1+x) ]dx
=-x[ ln(1-x) + ln(1+x) ] + ∫[ 1 -1/(1-x) +1 -1/(1+x) ]dx
=-x[ ln(1-x) + ln(1+x) ] + ∫[ 2 -1/(1-x) -1/(1+x) ]dx
=-x[ ln(1-x) + ln(1+x) ] + 2x +ln[(1-x)/(1+x)] + C
∫(0->1) ln[1/(1-x^2)] dx
=[ -x[ ln(1-x) + ln(1+x) ] + 2x +ln[(1-x)/(1+x)] ] |(0->1)
=ln2 + 2 + lim(x->1) { ln[(1-x)/(1+x)] - xln(1-x) }
=2+ln2 -ln2 - lim(x->1) (1-x)ln(1-x)
=2

追问

能写纸上传上来吗？

这样看好累