A wonderful Logarithmic integral solved using Feynman's technique

Integrate x ln(1 - x^2)/(1 + x^2)dx from 0 to 1 (Mis-1953)
#calculus #definite_integrals #feynmantechnique #maclaurinseries #substitution #cipher
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[@Mathematician6124]

It was intgrl x ln^2 (1-x^2) /(1+x^2) dx from 0 to 1. You forgot the square 😅

[@cipherunity]

It was too late when I found out. Consider it a new problem.

[@Mathematician6124]

@@cipherunity I'm writing here because that has 17 replies under that. OK here I write again. 2v ln((1-v)/(1+v)) ln(1+v) /(1+v^2) I just did 2v( ln(1-v) - ln(1+v)) ln(1+v) /(1+v^2) So, 2vln(1-v)ln(1+v) /(1+v^2) - 2vln(1+v) /(1+v^2) Then for, 2vln(1-v)ln(1+v) /(1+v^2) Focus on 2 ln(1-v) ln(1+v) only I used 4xy=(x+y)^2 - (x-y)^2 1/2 (4 ln(1-v)ln(1+v)) 1/2[ {(ln(1-v) + ln(1+v)} ^2 - {ln(1-v) - ln(1+v)} ^2] 1/2 ( ln^2 (1-v^2) - ln^2 (1-v /1+v)) Let me know.