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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)(tanaβˆ’1)  =>dx=(1/2)(1+tan^2 a)da     qd:   { ((xβˆ’>0)),((xβˆ’>∞)) :}=> { ((aβˆ’>(𝛑/4))),((aβˆ’>(𝛑/2))) :}  =>𝛀=2∫_(Ο€/4) ^(Ο€/2) ln^2 ((1/2)(tanaβˆ’1))dx  =>𝛀=2∫_(𝛑/4) ^(𝛑/2) ln^2 ((1/2))da+2∫_(Ο€/4) ^(Ο€/2) ln^2 (tanaβˆ’1)da  =>𝛀=(𝛑/2)ln^2 (2)+2∫_(𝛑/4) ^(𝛑/2) ln^2 (tanaβˆ’1)da  on  a I=2∫_(𝛑/4) ^(𝛑/2) ln^2 (tanaβˆ’1)da  posons tana=x=>da=(dx/(1+x^2 ))  =>I=2∫_1 ^∞ ((ln^2 (xβˆ’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.........

$$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{64}}\left(\boldsymbol{{a}\zeta}\left(\mathrm{2}\right)+{b}\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{2}\left[\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right]}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\Omega}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{x}}\right)}{\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posons}}\:\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1}=\boldsymbol{\mathrm{tana}}=>\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{tana}}βˆ’\mathrm{1}\right) \\ $$$$=>\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{da}}\:\:\:\:\:\boldsymbol{\mathrm{qd}}: \\ $$$$\begin{cases}{\boldsymbol{\mathrm{x}}βˆ’>\mathrm{0}}\\{\boldsymbol{\mathrm{x}}βˆ’>\infty}\end{cases}=>\begin{cases}{\boldsymbol{{a}}βˆ’>\frac{\boldsymbol{\pi}}{\mathrm{4}}}\\{\boldsymbol{{a}}βˆ’>\frac{\boldsymbol{\pi}}{\mathrm{2}}}\end{cases} \\ $$$$=>\boldsymbol{\Omega}=\mathrm{2}\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{2}} \boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{tana}}βˆ’\mathrm{1}\right)\right)\boldsymbol{\mathrm{dx}} \\ $$$$=>\boldsymbol{\Omega}=\mathrm{2}\int_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\boldsymbol{\mathrm{da}}+\mathrm{2}\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \left(\boldsymbol{\mathrm{tana}}βˆ’\mathrm{1}\right)\boldsymbol{\mathrm{da}} \\ $$$$=>\boldsymbol{\Omega}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{l}\overset{\mathrm{2}} {\mathrm{n}}\left(\mathrm{2}\right)+\mathrm{2}\int_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{tana}}βˆ’\mathrm{1}\right)\boldsymbol{\mathrm{da}} \\ $$$$\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{I}}=\mathrm{2}\int_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{tana}}βˆ’\mathrm{1}\right)\boldsymbol{\mathrm{da}} \\ $$$${posons}\:\boldsymbol{{tana}}=\boldsymbol{{x}}=>\boldsymbol{{da}}=\frac{\boldsymbol{{dx}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$=>\boldsymbol{\mathrm{I}}=\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{x}}βˆ’\mathrm{1}\right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }.\boldsymbol{\zeta}\left(\mathrm{2}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{I}}=\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)\int_{\mathrm{1}} ^{\infty} \frac{\boldsymbol{\mathrm{dx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)\left[\boldsymbol{{a}}\right]_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{I}}=\frac{\boldsymbol{\pi\zeta}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$=>\boldsymbol{\Omega}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left(\boldsymbol{\zeta}\left(\mathrm{2}\right)+\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\mathrm{2}\right)\right) \\ $$$$=>\boldsymbol{\Omega}=\frac{\boldsymbol{\pi}}{\mathrm{64}}\left(\mathrm{32}\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\mathrm{2}\right)+\mathrm{32}\boldsymbol{\zeta}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$$$...............{le}\:{celebre}\:{cedric}\:{junior}......... \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 14/Oct/22

thanks alot

$${thanks}\:{alot} \\ $$

Commented by Tawa11 last updated on 14/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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