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Question Number 140866 by liberty last updated on 13/May/21
∫10ln2−ln(1+x2)1−xdx=?
Answered by mindispower last updated on 15/May/21
=limx→1∫0xln(2)−ln(1+t2)1−tdt=limx→1−ln(2)ln(1−x)+ln(1−x)ln(1+x2)−∫0xln(1−t)2t1+t2dt=limx→1ln(1−x).ln(1+x22)=limx→1ln(1−1−x22)ln(1−x)=0=−∫01ln(1−x).2xdx1+x2=−AletB=∫01ln(1+x).2xdx1+x2A+B=∫01ln(1−x2)1+x2.d(x2)=∫01ln(1−t)1+tdt=∫01ln(t)2−tdt=∫012ln(2w)1−w.dw=ln2(2)+∫012ln(w)1−wdw=ln2(2)+∫112−ln(1−t)t=ln2(2)+li2(12)−li2(1)A−B=∫01ln(1−t1+t).2tdt1+t2,t=1−x1+xdt=−2(1+x)2dx=4∫01ln(x)2+2x2.1−x1+xdx∫01(1−x)ln(x)(1+x)(1+x2)dx=∫01ln(x)1+x−x1+x2ln(x)dx=34∫01ln(x)1+xdx=34.−∫01ln(1−(−x))−x.d(−x)=−∫0−1ln(1−t)tdt=−34li2(−1)
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