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Question Number 153896 by SANOGO last updated on 11/Sep/21

	Answered by ARUNG_Brandon_MBU last updated on 12/Sep/21
	$Ω=∫_1 ^∞ ((lnx)/((1+x)(1+x^2 )))dx, x=(1/u)⇒dx=−(1/u^2 )du =−∫_0 ^1 ((ulnu)/((u+1)(u^2 +1)))du (u/((u+1)(u^2 +1)))=(a/(u+1))+((bu+c)/(u^2 +1))=((a(u^2 +1)+(bu+c)(u+1))/((u+1)(u^2 +1))) { ((a+b=0)),((b+c=1)),((a+c=0)) :}⇒a=−(1/2), b=(1/2), c=(1/2) Ω=(1/2)∫_0 ^1 ((1/(u+1))−(u/(u^2 +1))−(1/(u^2 +1)))lnudu =(1/2)[∫_0 ^1 ((lnu)/(u+1))du−(1/4)∫_0 ^1 ((lnv)/(v+1))dv−(1/4)∫_0 ^1 ((v^(−(1/2)) lnv)/(v+1))dv] =(1/2)∫_0 ^1 (((1−u)lnu)/(1−u^2 ))du−(1/8)∫_0 ^1 (((1−v)lnv)/(1−v^2 ))dv−(1/8)∫_0 ^1 ((v^(−(1/2)) (1−v)lnv)/(1−v^2 ))dv =(1/8)∫_0 ^1 (((t^(−(1/2)) −1)lnt)/(1−t))dt−(1/(32))∫_0 ^1 (((t^(−(1/2)) −1)lnt)/(1−t))dt−(1/(32))∫_0 ^1 (((t^(−(1/4)) −t^(1/4) )lnt)/(1−t))dt =(1/8)(ψ′(1)−ψ′((1/2)))−(1/(32))(ψ′(1)−ψ′((1/2)))−(1/(32))(ψ′((5/4))−ψ′((3/4))) =(1/8)(ζ(2)−3ζ(2))−(1/(32))(ζ(2)−3ζ(2))−(1/(32))(ψ′((1/4))−ψ′((3/4))−16) =−(3/(16))ζ(2)−(1/(32))(2ψ′((1/4))−π^2 cosec^2 ((π/4)))+(1/2)$
	$Ω = \int_{1}^{\infty} \frac{\ln x}{(1 + x) (1 + x^{2})} d x, x = \frac{1}{u} \Rightarrow d x = - \frac{1}{u^{2}} d u$ $= - \int_{0}^{1} \frac{u \ln u}{(u + 1) (u^{2} + 1)} d u$ $\frac{u}{(u + 1) (u^{2} + 1)} = \frac{a}{u + 1} + \frac{b u + c}{u^{2} + 1} = \frac{a (u^{2} + 1) + (b u + c) (u + 1)}{(u + 1) (u^{2} + 1)}$ ${\begin{cases} a + b = 0 \\ b + c = 1 \\ a + c = 0 \end{cases} \Rightarrow a = - \frac{1}{2}, b = \frac{1}{2}, c = \frac{1}{2}$ $Ω = \frac{1}{2} \int_{0}^{1} (\frac{1}{u + 1} - \frac{u}{u^{2} + 1} - \frac{1}{u^{2} + 1}) \ln u d u$ $= \frac{1}{2} [\int_{0}^{1} \frac{\ln u}{u + 1} d u - \frac{1}{4} \int_{0}^{1} \frac{\ln v}{v + 1} d v - \frac{1}{4} \int_{0}^{1} \frac{v^{- \frac{1}{2}} \ln v}{v + 1} d v]$ $= \frac{1}{2} \int_{0}^{1} \frac{(1 - u) \ln u}{1 - u^{2}} d u - \frac{1}{8} \int_{0}^{1} \frac{(1 - v) \ln v}{1 - v^{2}} d v - \frac{1}{8} \int_{0}^{1} \frac{v^{- \frac{1}{2}} (1 - v) \ln v}{1 - v^{2}} d v$ $= \frac{1}{8} \int_{0}^{1} \frac{(t^{- \frac{1}{2}} - 1) \ln t}{1 - t} d t - \frac{1}{32} \int_{0}^{1} \frac{(t^{- \frac{1}{2}} - 1) \ln t}{1 - t} d t - \frac{1}{32} \int_{0}^{1} \frac{(t^{- \frac{1}{4}} - t^{\frac{1}{4}}) \ln t}{1 - t} d t$ $= \frac{1}{8} (ψ^{'} (1) - ψ^{'} (\frac{1}{2})) - \frac{1}{32} (ψ^{'} (1) - ψ^{'} (\frac{1}{2})) - \frac{1}{32} (ψ^{'} (\frac{5}{4}) - ψ^{'} (\frac{3}{4}))$ $= \frac{1}{8} (ζ (2) - 3 ζ (2)) - \frac{1}{32} (ζ (2) - 3 ζ (2)) - \frac{1}{32} (ψ^{'} (\frac{1}{4}) - ψ^{'} (\frac{3}{4}) - 16)$ $= - \frac{3}{16} ζ (2) - \frac{1}{32} (2 ψ^{'} (\frac{1}{4}) - π^{2} {cosec}^{2} (\frac{π}{4})) + \frac{1}{2}$
	Commented by SANOGO last updated on 12/Sep/21

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