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Question Number 72359 by aliesam last updated on 27/Oct/19

Commented by mind is power last updated on 27/Oct/19

nice one sir thanx

$$\mathrm{nice}\:\mathrm{one}\:\mathrm{sir}\:\mathrm{thanx} \\ $$

Commented by aliesam last updated on 27/Oct/19

thank you sir

$${thank}\:{you}\:{sir}\: \\ $$$$ \\ $$

Answered by mind is power last updated on 27/Oct/19

E=∫_0 ^(+∞) (((1+x+x^2 )log(x))/(1+x+x^2 +x^3 +x^4 ))dx+∫_0 ^(+∞) ((xlog(2)+x^2 log(3))/(1+x^2 +x^3 +x+x^4 ))dx 1+x+x^2 +x^3 +x^4 =(x^2 −2cos(((2π)/5))x+1)(x^2 −2cos(((4π)/5))x+1) cos(((2π)/5))=((−1+(√5))/4) cos(((4π)/5))=((2.(6−2(√5)))/(16))−1=((−4−4(√5))/(16))=−(((√5)+1)/4) 1+x+x^2 +x^3 +x^4 =(x^2 −(((−1+(√5))/2))x+1)(x^2 +(((1+(√5))/2))x+1) ∫_0 ^(+∞) (((1+x+x^2 )log(x))/(x^4 +x^3 +x^2 +x+1))dx let t=(1/x) ∫_0 ^(+∞) (((1+x+x^2 )log(x))/(x^4 +x^3 +x^2 +x+1))dx=∫_∞ ^0 (((1+(1/t)+(1/t^2 ))(−log(t)))/(((1/t^4 )+(1/t^3 )+(1/t^2 )+(1/t^ )+1))).((−dt)/( t^2 )) =∫_∞ ^0 (((t^2 +t+1)log(t))/((1+t+t^2 +t^3 +t^4 )))=−∫_0 ^(+∞) (((1+x+x^2 )log(x))/(x^4 +x^3 +x^2 +x+1))dx⇒ 2∫_0 ^(+∞) (((1+x+x^2 )log(x))/(x^4 +x^3 +x^2 +x+1))dx=0⇒∫_0 ^(+∞) (((1+x+x^2 )log(x))/(x^4 +x^3 +x^2 +x+1))dx=0 ⇔ E=∫_0 ^(+∞) ((xlog(2)+x^2 log(3))/(1+x+x^2 +x^3 +x^4 ))=∫_0 ^(+∞) ((xlog(2)+x^2 log(3))/((x^2 −(((−1+(√5))/2))x+1)(x^2 +(((1+(√5))/2))x+1))) ((ax+b)/(x^2 −(((−1+(√5))/2))x+1))+((cx+d)/((x^2 +(((1+(√5))/2))x+1)) a+c=0 b+d=0 (a−c)((√5)/2)=log(3) (b−d)((√5)/2)=log(2) a=((log(3))/(√5)),c=−((log(3))/(√5)) b=((log(2))/(√5)),d=((−log(2))/(√5)) ((ax+b)/(x^2 −(((−1+(√5))/2))x+1))+((cx+d)/((x^2 +(((1+(√5))/2))x+1))=(1/(√5))(((xlog(3)+log(2))/(x^2 −(((−1+(√5))/2))x+1))+((−xlog(3)−log(2))/(x^2 +(((1+(√5))/2))x+1))) now just integrat bee continued

$$ \\ $$$$\mathrm{E}=\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} }\mathrm{dx}+\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{xlog}\left(\mathrm{2}\right)+\mathrm{x}^{\mathrm{2}} \mathrm{log}\left(\mathrm{3}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} =\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)=\frac{\mathrm{2}.\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\right)}{\mathrm{16}}−\mathrm{1}=\frac{−\mathrm{4}−\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{16}}=−\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} =\left(\mathrm{x}^{\mathrm{2}} −\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mathrm{dx}=\int_{\infty} ^{\mathrm{0}} \frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\left(−\mathrm{log}\left(\mathrm{t}\right)\right)}{\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{t}^{} }+\mathrm{1}\right)}.\frac{−\mathrm{dt}}{\:\mathrm{t}^{\mathrm{2}} } \\ $$$$=\int_{\infty} ^{\mathrm{0}} \frac{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}\right)\mathrm{log}\left(\mathrm{t}\right)}{\left(\mathrm{1}+\mathrm{t}+\mathrm{t}^{\mathrm{2}} +\mathrm{t}^{\mathrm{3}} +\mathrm{t}^{\mathrm{4}} \right)}=−\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mathrm{dx}\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mathrm{dx}=\mathrm{0}\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mathrm{dx}=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{E}=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{xlog}\left(\mathrm{2}\right)+\mathrm{x}^{\mathrm{2}} \mathrm{log}\left(\mathrm{3}\right)}{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} }=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{xlog}\left(\mathrm{2}\right)+\mathrm{x}^{\mathrm{2}} \mathrm{log}\left(\mathrm{3}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} −\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}}+\frac{\mathrm{cx}+\mathrm{d}}{\left(\mathrm{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}\right.} \\ $$$$\mathrm{a}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{b}+\mathrm{d}=\mathrm{0} \\ $$$$\left(\mathrm{a}−\mathrm{c}\right)\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{log}\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{b}−\mathrm{d}\right)\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{log}\left(\mathrm{2}\right) \\ $$$$\mathrm{a}=\frac{\mathrm{log}\left(\mathrm{3}\right)}{\sqrt{\mathrm{5}}},\mathrm{c}=−\frac{\mathrm{log}\left(\mathrm{3}\right)}{\sqrt{\mathrm{5}}} \\ $$$$\mathrm{b}=\frac{\mathrm{log}\left(\mathrm{2}\right)}{\sqrt{\mathrm{5}}},\mathrm{d}=\frac{−\mathrm{log}\left(\mathrm{2}\right)}{\sqrt{\mathrm{5}}} \\ $$$$\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} −\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}}+\frac{\mathrm{cx}+\mathrm{d}}{\left(\mathrm{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}\right.}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\frac{\mathrm{xlog}\left(\mathrm{3}\right)+\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{x}^{\mathrm{2}} −\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}}+\frac{−\mathrm{xlog}\left(\mathrm{3}\right)−\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{x}+\mathrm{1}}\right) \\ $$$$\mathrm{now}\:\mathrm{just}\:\mathrm{integrat}\:\mathrm{bee}\:\mathrm{continued} \\ $$$$ \\ $$$$ \\ $$

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