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Question Number 178246 by mnjuly1970 last updated on 14/Oct/22

	Answered by CElcedricjunior last updated on 14/Oct/22
	$𝛀=∫_0 ^∞ ((ln^2 (x))/(1+2x+2x^2 ))dx=(𝛑/(64))(a𝛇(2)+bln^2 (2)) 𝛀=∫_0 ^∞ ((ln^2 (x))/(1+2x+2x^2 ))=∫_0 ^∞ ((ln^2 (x))/(2[(x+(1/2))^2 +(1/4)]))dx 𝛀=2∫_0 ^∞ ((ln^2 (x))/((2x+1)^2 +1))dx posons 2x+1=tana=>x=(1/2)(tana−1) =>dx=(1/2)(1+tan^2 a)da qd: { ((x−>0)),((x−>∞)) :}=> { ((a−>(𝛑/4))),((a−>(𝛑/2))) :} =>𝛀=2∫_(π/4) ^(π/2) ln^2 ((1/2)(tana−1))dx =>𝛀=2∫_(𝛑/4) ^(𝛑/2) ln^2 ((1/2))da+2∫_(π/4) ^(π/2) ln^2 (tana−1)da =>𝛀=(𝛑/2)ln^2 (2)+2∫_(𝛑/4) ^(𝛑/2) ln^2 (tana−1)da on a I=2∫_(𝛑/4) ^(𝛑/2) ln^2 (tana−1)da posons tana=x=>da=(dx/(1+x^2 )) =>I=2∫_1 ^∞ ((ln^2 (x−1))/(1+x^2 ))dx=2∫_1 ^∞ (1/(1+x^2 )).𝛇(2)dx I=2𝛇(2)∫_1 ^∞ (dx/(1+x^2 ))=2𝛇(2)[a]_(𝛑/4) ^(𝛑/2) I=((𝛑𝛇(2))/2) =>𝛀=(𝛑/2)(𝛇(2)+ln^2 (2)) =>𝛀=(𝛑/(64))(32ln^2 (2)+32𝛇(2)) ...............le celebre cedric junior.........$
	$Ω = \int_{0}^{\infty} \frac{\ln^{2} (x)}{1 + 2 x + 2 x^{2}} dx = \frac{π}{64} (a ζ (2) + b \ln^{2} (2))$ $Ω = \int_{0}^{\infty} \frac{\ln^{2} (x)}{1 + 2 x + 2 x^{2}} = \int_{0}^{\infty} \frac{l \overset{2}{n} (x)}{2 [{(x + \frac{1}{2})}^{2} + \frac{1}{4}]} dx$ $Ω = 2 \int_{0}^{\infty} \frac{l \overset{2}{n} (x)}{{(2 x + 1)}^{2} + 1} dx$ $posons 2 x + 1 = tana => x = \frac{1}{2} (tana - 1)$ $=> dx = \frac{1}{2} (1 + \tan^{2} a) da qd :$ ${\begin{cases} x - > 0 \\ x - > \infty \end{cases} => {\begin{cases} a - > \frac{π}{4} \\ a - > \frac{π}{2} \end{cases}$ $=> Ω = 2 \int_{π / 4}^{π / 2} \ln^{2} (\frac{1}{2} (tana - 1)) dx$ $=> Ω = 2 \int_{π / 4}^{π / 2} \ln^{2} (\frac{1}{2}) da + 2 \int_{π / 4}^{π / 2} \ln^{2} (tana - 1) da$ $=> Ω = \frac{π}{2} l \overset{2}{n} (2) + 2 \int_{π / 4}^{π / 2} l \overset{2}{n} (tana - 1) da$ $on a I = 2 \int_{π / 4}^{π / 2} \ln^{2} (tana - 1) da$ $p o s o n s t a n a = x => d a = \frac{d x}{1 + x^{2}}$ $=> I = 2 \int_{1}^{\infty} \frac{l \overset{2}{n} (x - 1)}{1 + x^{2}} dx = 2 \int_{1}^{\infty} \frac{1}{1 + x^{2}} . ζ (2) dx$ $I = 2 ζ (2) \int_{1}^{\infty} \frac{dx}{1 + x^{2}} = 2 ζ (2) {[a]}_{π / 4}^{π / 2}$ $I = \frac{π ζ (2)}{2}$ $=> Ω = \frac{π}{2} (ζ (2) + l \overset{2}{n} (2))$ $=> Ω = \frac{π}{64} (32 l \overset{2}{n} (2) + 32 ζ (2))$ $. . . . . . . . . . . . . . . l e c e l e b r e c e d r i c j u n i o r . . . . . . . . .$
	Commented by mnjuly1970 last updated on 14/Oct/22

	$t h a n k s a l o t$
	Commented by Tawa11 last updated on 14/Oct/22

	$Great sir$
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