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Question Number 64662 by aliesam last updated on 20/Jul/19

∫(dx/((x^8 +x^4 +1)^2 )) ∫_(1/x) ^x ((ln(t))/(t^2 +1)) dt

$$\int\frac{{dx}}{\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}\:{dt} \\ $$

Commented by MJS last updated on 20/Jul/19

the 2^(nd) one should be =0

$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{should}\:\mathrm{be}\:=\mathrm{0} \\ $$

Commented by aliesam last updated on 20/Jul/19

yes thats right

$${yes}\:{thats}\:{right} \\ $$

Commented by mathmax by abdo last updated on 20/Jul/19

2) let I =∫_(1/x) ^x ((ln(t))/(1+t^2 ))dt with x>0 changement t =(1/u) give I =−∫_(1/x) ^x ((−lnu)/(1+(1/u^2 )))(−(du/u^2 )) =−∫_(1/x) ^x ((lnu)/(u^2 +1)) =−I ⇒2I=0 ⇒I =0 remark when x go to +∞ we get ∫_0 ^(+∞) ((lnt)/(t^2 +1)) dt =0

$$\left.\mathrm{2}\right)\:{let}\:\:{I}\:=\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0}\:\:{changement}\:{t}\:=\frac{\mathrm{1}}{{u}}\:{give} \\ $$$${I}\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\:\frac{−{lnu}}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right)\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\:\frac{{lnu}}{{u}^{\mathrm{2}} +\mathrm{1}}\:=−{I}\:\Rightarrow\mathrm{2}{I}=\mathrm{0}\:\Rightarrow{I}\:=\mathrm{0} \\ $$$${remark}\:{when}\:{x}\:{go}\:{to}\:+\infty\:\:{we}\:{get}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{lnt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:=\mathrm{0} \\ $$

Answered by MJS last updated on 20/Jul/19

∫(dx/((x^8 +x^4 +1)^2 )) Ostrogradski′s Method ∫((P(x))/(Q(x)))dx=((P_1 (x))/(Q_1 (x)))+∫((P_2 (x))/(Q_2 (x)))dx Q_1 (x)=gcd (Q(x), Q′(x)) Q_2 (x)=((Q(x))/(Q_1 (x))) P_1 (x), P_2 (x) can be found by comparing the factors of ((P(x))/(Q(x)))=(((P_1 (x))/(Q_1 (x))))′+((P_2 (x))/(Q_2 (x))) in our case P(x)=1 Q(x)=(x^8 +x^4 +1)^2 Q_1 (x)=Q_2 (x)=x^8 +x^4 +1 P_1 (x)=(1/(12))(x−x^5 ) P_2 (x)=(1/(12))(11−3x^4 ) ∫(dx/((x^8 +x^4 +1)^2 ))=−((x^5 −x)/(12(x^8 +x^4 +1)))−(1/(12))∫((3x^4 −11)/(x^8 +x^4 +1))dx ((3x^4 −11)/(x^8 +x^4 +1))=((3x^4 −11)/((x^2 −x+1)(x^2 +x+1)(x^2 −(√3)x+1)(x^2 +(√3)x+1)))= =−((3x+11)/(4(x^2 −x+1)))+((3x−11)/(4(x^2 +x+1)))+((25x−11(√3))/(4(√3)(x^2 −(√3)x+1)))−((25x+11(√3))/(4(√3)(x^2 +(√3)x+1))) now use this formula ∫((ax+b)/(x^2 +cx+d))dx=(a/2)∫((2x+c)/(x^2 +cx+d))dx+((2b−ac)/2)∫(dx/(x^2 +cx+d))= =(a/2)ln (x^2 +cx+d) −((ac−2b)/(√(4d−c^2 )))arctan ((2x+c)/(√(4d−c^2 )))

$$\int\frac{{dx}}{\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method} \\ $$$$\int\frac{{P}\left({x}\right)}{{Q}\left({x}\right)}{dx}=\frac{{P}_{\mathrm{1}} \left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)}+\int\frac{{P}_{\mathrm{2}} \left({x}\right)}{{Q}_{\mathrm{2}} \left({x}\right)}{dx} \\ $$$${Q}_{\mathrm{1}} \left({x}\right)=\mathrm{gcd}\:\left({Q}\left({x}\right),\:{Q}'\left({x}\right)\right) \\ $$$${Q}_{\mathrm{2}} \left({x}\right)=\frac{{Q}\left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)} \\ $$$${P}_{\mathrm{1}} \left({x}\right),\:{P}_{\mathrm{2}} \left({x}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{by}\:\mathrm{comparing}\:\mathrm{the} \\ $$$$\mathrm{factors}\:\mathrm{of}\:\frac{{P}\left({x}\right)}{{Q}\left({x}\right)}=\left(\frac{{P}_{\mathrm{1}} \left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)}\right)'+\frac{{P}_{\mathrm{2}} \left({x}\right)}{{Q}_{\mathrm{2}} \left({x}\right)} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$${P}\left({x}\right)=\mathrm{1}\:\:{Q}\left({x}\right)=\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$${Q}_{\mathrm{1}} \left({x}\right)={Q}_{\mathrm{2}} \left({x}\right)={x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1} \\ $$$${P}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{12}}\left({x}−{x}^{\mathrm{5}} \right)\:\:{P}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{11}−\mathrm{3}{x}^{\mathrm{4}} \right) \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }=−\frac{{x}^{\mathrm{5}} −{x}}{\mathrm{12}\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{12}}\int\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{11}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$$$ \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{11}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{11}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}= \\ $$$$=−\frac{\mathrm{3}{x}+\mathrm{11}}{\mathrm{4}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}+\frac{\mathrm{3}{x}−\mathrm{11}}{\mathrm{4}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}+\frac{\mathrm{25}{x}−\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}−\frac{\mathrm{25}{x}+\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula} \\ $$$$\int\frac{{ax}+{b}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx}=\frac{{a}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{c}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx}+\frac{\mathrm{2}{b}−{ac}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +{cx}+{d}}= \\ $$$$=\frac{{a}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{cx}+{d}\right)\:−\frac{{ac}−\mathrm{2}{b}}{\sqrt{\mathrm{4}{d}−{c}^{\mathrm{2}} }}\mathrm{arctan}\:\frac{\mathrm{2}{x}+{c}}{\sqrt{\mathrm{4}{d}−{c}^{\mathrm{2}} }} \\ $$

Commented by aliesam last updated on 20/Jul/19

god bless uou sir

$${god}\:{bless}\:{uou}\:{sir}\: \\ $$

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