Question Number 64662 by aliesam last updated on 20/Jul/19
$$\int\frac{{dx}}{\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}\:{dt} \\ $$
Commented by MJS last updated on 20/Jul/19
$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{should}\:\mathrm{be}\:=\mathrm{0} \\ $$
Commented by aliesam last updated on 20/Jul/19
$${yes}\:{thats}\:{right} \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
$$\left.\mathrm{2}\right)\:{let}\:\:{I}\:=\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0}\:\:{changement}\:{t}\:=\frac{\mathrm{1}}{{u}}\:{give} \\ $$$${I}\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\:\frac{−{lnu}}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right)\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\:\frac{{lnu}}{{u}^{\mathrm{2}} +\mathrm{1}}\:=−{I}\:\Rightarrow\mathrm{2}{I}=\mathrm{0}\:\Rightarrow{I}\:=\mathrm{0} \\ $$$${remark}\:{when}\:{x}\:{go}\:{to}\:+\infty\:\:{we}\:{get}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{lnt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:=\mathrm{0} \\ $$
Answered by MJS last updated on 20/Jul/19
$$\int\frac{{dx}}{\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method} \\ $$$$\int\frac{{P}\left({x}\right)}{{Q}\left({x}\right)}{dx}=\frac{{P}_{\mathrm{1}} \left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)}+\int\frac{{P}_{\mathrm{2}} \left({x}\right)}{{Q}_{\mathrm{2}} \left({x}\right)}{dx} \\ $$$${Q}_{\mathrm{1}} \left({x}\right)=\mathrm{gcd}\:\left({Q}\left({x}\right),\:{Q}'\left({x}\right)\right) \\ $$$${Q}_{\mathrm{2}} \left({x}\right)=\frac{{Q}\left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)} \\ $$$${P}_{\mathrm{1}} \left({x}\right),\:{P}_{\mathrm{2}} \left({x}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{by}\:\mathrm{comparing}\:\mathrm{the} \\ $$$$\mathrm{factors}\:\mathrm{of}\:\frac{{P}\left({x}\right)}{{Q}\left({x}\right)}=\left(\frac{{P}_{\mathrm{1}} \left({x}\right)}{{Q}_{\mathrm{1}} \left({x}\right)}\right)'+\frac{{P}_{\mathrm{2}} \left({x}\right)}{{Q}_{\mathrm{2}} \left({x}\right)} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$${P}\left({x}\right)=\mathrm{1}\:\:{Q}\left({x}\right)=\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$${Q}_{\mathrm{1}} \left({x}\right)={Q}_{\mathrm{2}} \left({x}\right)={x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1} \\ $$$${P}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{12}}\left({x}−{x}^{\mathrm{5}} \right)\:\:{P}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{11}−\mathrm{3}{x}^{\mathrm{4}} \right) \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }=−\frac{{x}^{\mathrm{5}} −{x}}{\mathrm{12}\left({x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{12}}\int\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{11}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$$$ \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{11}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{11}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}= \\ $$$$=−\frac{\mathrm{3}{x}+\mathrm{11}}{\mathrm{4}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}+\frac{\mathrm{3}{x}−\mathrm{11}}{\mathrm{4}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}+\frac{\mathrm{25}{x}−\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}−\frac{\mathrm{25}{x}+\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula} \\ $$$$\int\frac{{ax}+{b}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx}=\frac{{a}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{c}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx}+\frac{\mathrm{2}{b}−{ac}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +{cx}+{d}}= \\ $$$$=\frac{{a}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{cx}+{d}\right)\:−\frac{{ac}−\mathrm{2}{b}}{\:\sqrt{\mathrm{4}{d}−{c}^{\mathrm{2}} }}\mathrm{arctan}\:\frac{\mathrm{2}{x}+{c}}{\:\sqrt{\mathrm{4}{d}−{c}^{\mathrm{2}} }} \\ $$
Commented by aliesam last updated on 20/Jul/19
$${god}\:{bless}\:{uou}\:{sir}\: \\ $$