Question Number 178246 by mnjuly1970 last updated on 14/Oct/22
Answered by CElcedricjunior last updated on 14/Oct/22
![π=β«_0 ^β ((ln^2 (x))/(1+2x+2x^2 ))dx=(π/(64))(aπ(2)+bln^2 (2)) π=β«_0 ^β ((ln^2 (x))/(1+2x+2x^2 ))=β«_0 ^β ((ln^2 (x))/(2[(x+(1/2))^2 +(1/4)]))dx π=2β«_0 ^β ((ln^2 (x))/((2x+1)^2 +1))dx posons 2x+1=tana=>x=(1/2)(tanaβ1) =>dx=(1/2)(1+tan^2 a)da qd: { ((xβ>0)),((xβ>β)) :}=> { ((aβ>(π/4))),((aβ>(π/2))) :} =>π=2β«_(Ο/4) ^(Ο/2) ln^2 ((1/2)(tanaβ1))dx =>π=2β«_(π/4) ^(π/2) ln^2 ((1/2))da+2β«_(Ο/4) ^(Ο/2) ln^2 (tanaβ1)da =>π=(π/2)ln^2 (2)+2β«_(π/4) ^(π/2) ln^2 (tanaβ1)da on a I=2β«_(π/4) ^(π/2) ln^2 (tanaβ1)da posons tana=x=>da=(dx/(1+x^2 )) =>I=2β«_1 ^β ((ln^2 (xβ1))/(1+x^2 ))dx=2β«_1 ^β (1/(1+x^2 )).π(2)dx I=2π(2)β«_1 ^β (dx/(1+x^2 ))=2π(2)[a]_(π/4) ^(π/2) I=((ππ(2))/2) =>π=(π/2)(π(2)+ln^2 (2)) =>π=(π/(64))(32ln^2 (2)+32π(2)) ...............le celebre cedric junior.........](https://www.tinkutara.com/question/Q178285.png)
$$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{64}}\left(\boldsymbol{{a}\zeta}\left(\mathrm{2}\right)+{b}\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{2}\left[\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right]}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\Omega}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{x}}\right)}{\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posons}}\:\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1}=\boldsymbol{\mathrm{tana}}=>\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{tana}}β\mathrm{1}\right) \\ $$$$=>\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\mathrm{a}}\right)\boldsymbol{\mathrm{da}}\:\:\:\:\:\boldsymbol{\mathrm{qd}}: \\ $$$$\begin{cases}{\boldsymbol{\mathrm{x}}β>\mathrm{0}}\\{\boldsymbol{\mathrm{x}}β>\infty}\end{cases}=>\begin{cases}{\boldsymbol{{a}}β>\frac{\boldsymbol{\pi}}{\mathrm{4}}}\\{\boldsymbol{{a}}β>\frac{\boldsymbol{\pi}}{\mathrm{2}}}\end{cases} \\ $$$$=>\boldsymbol{\Omega}=\mathrm{2}\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{2}} \boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{tana}}β\mathrm{1}\right)\right)\boldsymbol{\mathrm{dx}} \\ $$$$=>\boldsymbol{\Omega}=\mathrm{2}\int_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\boldsymbol{\mathrm{da}}+\mathrm{2}\int_{\pi/\mathrm{4}} ^{\pi/\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \left(\boldsymbol{\mathrm{tana}}β\mathrm{1}\right)\boldsymbol{\mathrm{da}} \\ $$$$=>\boldsymbol{\Omega}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{l}\overset{\mathrm{2}} {\mathrm{n}}\left(\mathrm{2}\right)+\mathrm{2}\int_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{tana}}β\mathrm{1}\right)\boldsymbol{\mathrm{da}} \\ $$$$\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{I}}=\mathrm{2}\int_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{tana}}β\mathrm{1}\right)\boldsymbol{\mathrm{da}} \\ $$$${posons}\:\boldsymbol{{tana}}=\boldsymbol{{x}}=>\boldsymbol{{da}}=\frac{\boldsymbol{{dx}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$=>\boldsymbol{\mathrm{I}}=\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\boldsymbol{\mathrm{x}}β\mathrm{1}\right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }.\boldsymbol{\zeta}\left(\mathrm{2}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{I}}=\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)\int_{\mathrm{1}} ^{\infty} \frac{\boldsymbol{\mathrm{dx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)\left[\boldsymbol{{a}}\right]_{\boldsymbol{\pi}/\mathrm{4}} ^{\boldsymbol{\pi}/\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{I}}=\frac{\boldsymbol{\pi\zeta}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$=>\boldsymbol{\Omega}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\left(\boldsymbol{\zeta}\left(\mathrm{2}\right)+\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\mathrm{2}\right)\right) \\ $$$$=>\boldsymbol{\Omega}=\frac{\boldsymbol{\pi}}{\mathrm{64}}\left(\mathrm{32}\boldsymbol{\mathrm{l}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}}}\left(\mathrm{2}\right)+\mathrm{32}\boldsymbol{\zeta}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$$$……………{le}\:{celebre}\:{cedric}\:{junior}……… \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 14/Oct/22
$${thanks}\:{alot} \\ $$
Commented by Tawa11 last updated on 14/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$