Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 136036 by mathmax by abdo last updated on 18/Mar/21

calculate ∫_0 ^∞ ((logx)/(x^4 +x^2 +1))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$

Commented by Ajetunmobi last updated on 18/Mar/21

hi sir how can we chat privately ?

$$ \\ $$$$\boldsymbol{\mathrm{hi}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{chat}}\:\boldsymbol{\mathrm{privately}}\:? \\ $$

Commented by Ajetunmobi last updated on 18/Mar/21

Commented by liki last updated on 19/Mar/21

Drop your whatssap number here !

$$\mathrm{Drop}\:\mathrm{your}\:\mathrm{whatssap}\:\mathrm{number}\:\mathrm{here}\:! \\ $$

Commented by mathmax by abdo last updated on 31/May/21

you are right liki...

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{liki}... \\ $$

Answered by Ar Brandon last updated on 18/Mar/21

Δ=∫_0 ^∞ ((lnx)/(x^4 +x^2 +1))dx =∫_0 ^1 ((lnx)/(x^4 +x^2 +1))dx+∫_1 ^∞ ((lnx)/(x^4 +x^2 +1))dx =∫_0 ^1 ((lnx)/(x^4 +x^2 +1))dx−∫_0 ^1 ((x^2 lnx)/(x^4 +x^2 +1))dx =∫_0 ^1 (((1−x^2 )lnx)/((1−x^6 )))dx−∫_0 ^1 ((x^2 (1−x^2 )lnx)/((1−x^6 )))dx =(1/(36))∫_0 ^1 (((u^(−(5/6)) −u^(−(1/2)) )lnu)/(1−u))du−(1/(36))∫_0 ^1 (((u^(−(1/2)) −u^(−(1/6)) )lnu)/(1−u))du =−(1/(36)){ψ^1 ((1/6))−ψ^1 ((1/2))−ψ^1 ((1/2))+ψ^1 ((5/6))} =−(1/(36)){(π^2 /(sin^2 ((π/6))))−2ψ^1 ((1/2))}=−(1/(36)){4π^2 −2×4×(3/4)ζ(2)} =−(((4π^2 −π^2 ))/(36))=−(π^2 /(12))★

$$\Delta=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{lnx}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{6}} \right)}\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{lnx}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{6}} \right)}\mathrm{dx} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{36}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{u}^{−\frac{\mathrm{5}}{\mathrm{6}}} −\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)\mathrm{lnu}}{\mathrm{1}−\mathrm{u}}\mathrm{du}−\frac{\mathrm{1}}{\mathrm{36}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{6}}} \right)\mathrm{lnu}}{\mathrm{1}−\mathrm{u}}\mathrm{du} \\ $$$$\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{36}}\left\{\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)−\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\psi^{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right\} \\ $$$$\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{36}}\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)}−\mathrm{2}\psi^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}=−\frac{\mathrm{1}}{\mathrm{36}}\left\{\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)\right\} \\ $$$$\:\:\:\:\:=−\frac{\left(\mathrm{4}\pi^{\mathrm{2}} −\pi^{\mathrm{2}} \right)}{\mathrm{36}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\bigstar \\ $$

Commented by Dwaipayan Shikari last updated on 18/Mar/21

😃

$$ \\ $$😃

Commented by Ajetunmobi last updated on 18/Mar/21

you did the same solution as mine but why?

$$ \\ $$$$\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{did}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{mine}}\: \\ $$$$\boldsymbol{\mathrm{but}}\:\boldsymbol{\mathrm{why}}? \\ $$

Commented by Ar Brandon last updated on 18/Mar/21

Because I wanna learn from you too. Just as I do with the other members. 😄😇

$$\mathrm{Because}\:\mathrm{I}\:\mathrm{wanna}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{you}\:\mathrm{too}. \\ $$$$\mathrm{Just}\:\mathrm{as}\:\mathrm{I}\:\mathrm{do}\:\mathrm{with}\:\mathrm{the}\:\mathrm{other}\:\mathrm{members}. \\ $$😄😇

Commented by Ajetunmobi last updated on 18/Mar/21

ok sir we are all learnig from each other

$$ \\ $$$$\boldsymbol{\mathrm{ok}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{learnig}}\:\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{each}}\: \\ $$$$\boldsymbol{\mathrm{other}} \\ $$

Commented by Ar Brandon last updated on 18/Mar/21

😀

😀

Commented by SLVR last updated on 13/Apr/21

good evening mr.ar.Brandon kidly help me what is ψ′(1/6).. how it is

$${good}\:{evening}\:{mr}.{ar}.{Brandon}\: \\ $$$${kidly}\:{help}\:{me}\:{what}\:\:{is}\:\psi'\left(\mathrm{1}/\mathrm{6}\right).. \\ $$$${how}\:{it}\:{is} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com