I:=∫_0 ^( 1) ((ln (x))/(1 + x^( 2) )) dx := ∫_0 ^( 1) ln(x ) Σ_(n=0) ^∞ (−1)^( n) x^( 2n) dx := Σ_(n=0) ^∞ ( −1 )^( n) ∫_0 ^( 1) x^( 2n) ln( x )dx : = Σ_(n=0) ^∞ ( −1 )^( n) { [(x^( 2n+1) /(2n +1)) ln ( x )]_0 ^( 1) −(1/((2n +1 )^( 2) )) } : = Σ_(n=1) ^( ∞) ((( −1 )^( n−1) )/(( 2n +1)^( 2) )) = −G (Catalan constant )


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Question Number 144813 by mnjuly1970 last updated on 29/Jun/21
$I:=∫_0 ^( 1) ((ln (x))/(1 + x^( 2) )) dx := ∫_0 ^( 1) ln(x ) Σ_(n=0) ^∞ (−1)^( n) x^( 2n) dx := Σ_(n=0) ^∞ ( −1 )^( n) ∫_0 ^( 1) x^( 2n) ln( x )dx : = Σ_(n=0) ^∞ ( −1 )^( n) { [(x^( 2n+1) /(2n +1)) ln ( x )]_0 ^( 1) −(1/((2n +1 )^( 2) )) } : = Σ_(n=1) ^( ∞) ((( −1 )^( n−1) )/(( 2n +1)^( 2) )) = −G (Catalan constant )$
$I := \int_{0}^{1} \frac{\ln (x)}{1 + x^{2}} d x$ $:= \int_{0}^{1} \ln (x) \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{2 n} d x$ $:= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{2 n} \ln (x) d x$ $: = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {{[\frac{x^{2 n + 1}}{2 n + 1} \ln (x)]}_{0}^{1} - \frac{1}{{(2 n + 1)}^{2}}}$ $: = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{{(2 n + 1)}^{2}} = - G (Catalan constant)$
Commented by Dwaipayan Shikari last updated on 29/Jun/21

$\int_{0}^{1} \frac{l o g (x)}{1 + x^{2}} d x = \int_{0}^{1} \frac{1 - x^{2}}{1 - x^{4}} l o g (x) d x$ $= \frac{1}{16} \int_{0}^{1} \frac{u^{- \frac{3}{4}} - u^{- \frac{1}{4}}}{1 - u} l o g (u) d u$ $= \frac{1}{16} (ψ^{'} (\frac{1}{4}) - ψ^{'} (\frac{3}{4})) = - G$
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