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∫01ln(x)1−x2 dx\large \int_{0}^{1} \dfrac{\ln( x)}{1-x^2} \, dx ∫011−x2ln(x)dx
The above integral is equal to −aπbc\dfrac{-a\pi^b}{c}c−aπb for positive integers a,b,ca,b,ca,b,c, for positive integers a,ba,ba,b and ccc with aaa and ccc coprime. Evaluate (a+b+c)2(a+b+c)^2(a+b+c)2.
Note
You are given that ζ(2)=π26\zeta(2) = \dfrac{\pi^2}{6} ζ(2)=6π2.
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