Question Number 181403 by a.lgnaoui last updated on 24/Nov/22
![Refer to Q181319 ∫((x^4 +1)/(x^4 +x+2))dx ((x^4 +1)/(x^4 +x+2))=1+(x/(2(x^2 +x+1)))−(x/(2(x^2 −x+1))) =1+(1/4)[(((2x+1)−1)/(x^2 +x+1))−(1/4)×(((2x−1)+1)/((x^2 −x+1))) =(1/4)×[log(x^2 +x+1)]^′ −(1/4)×((1/([((√3)/2)(x+(1/2))^2 ]+1))) −((1/4)[log(x^2 −x+1)]^′ +(1/4)((1/([((√3)/2)×(x−(1/2)]^2 +1)))) first with [((√3)/2)( x+(1/2))]=u ∫(1/(u^2 +1))du=arctg(u) and v=[((√3)/2)(x−(1/2) )]=v ∫(dv/(1+v^2 ))=arctg(v) ...............](https://www.tinkutara.com/question/Q181403.png)
$${Refer}\:{to}\:\mathrm{Q181319} \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}+\mathrm{2}}{dx} \\ $$$$\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}+\mathrm{2}}=\mathrm{1}+\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\left[\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)\right]^{'} −\frac{\mathrm{1}}{\mathrm{4}}×\left(\frac{\mathrm{1}}{\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right]+\mathrm{1}}\right) \\ $$$$−\left(\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{log}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)\right]^{'} +\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right]^{\mathrm{2}} +\mathrm{1}\right.}\right)\right) \\ $$$${first}\:{with}\:\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right]={u}\:\:\:\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}={arctg}\left({u}\right) \\ $$$${and}\:{v}=\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\:\right)\right]=\mathrm{v}\:\:\:\int\frac{{dv}}{\mathrm{1}+{v}^{\mathrm{2}} }={arctg}\left({v}\right) \\ $$$$…………… \\ $$
Answered by ARUNG_Brandon_MBU last updated on 24/Nov/22
![I=∫((x^4 +1)/(x^4 +x^2 +1))dx=∫(((x^4 +x^2 +1)−x^2 )/(x^4 +x^2 +1))dx =∫(1−(x^2 /(x^4 +x^2 +1)))dx=x−∫(x^2 /(x^4 +x^2 +1))dx =x−(1/2)∫(((x^2 +1)+(x^2 −1))/(x^4 +x^2 +1))dx =x−(1/2)∫((x^2 +1)/(x^4 +x^2 +1))dx−(1/2)∫((x^2 −1)/(x^4 +x^2 +1))dx =x−(1/2)∫((1+(1/x^2 ))/(x^2 +1+(1/x^2 )))dx−(1/2)∫((1−(1/x^2 ))/(x^2 +1+(1/x^2 )))dx =x−(1/2)∫((1+(1/x^2 ))/((x−(1/x))^2 +3))dx−(1/2)∫((1−(1/x^2 ))/((x+(1/x))^2 −1))dx =x−(1/(2(√3)))arctan[(1/( (√3)))(x−(1/x))]+(1/2)arcoth(x+(1/x))+C =x−((√3)/6)arctan(((x^2 −1)/(x(√3))))+(1/4)ln∣((x^2 +x+1)/(x^2 −x+1))∣+C](https://www.tinkutara.com/question/Q181413.png)
$${I}=\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\int\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)−{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:=\int\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}={x}−\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:={x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:={x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:={x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\:\:={x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$\:\:={x}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\mathrm{arctan}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}−\frac{\mathrm{1}}{{x}}\right)\right]+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcoth}\left({x}+\frac{\mathrm{1}}{{x}}\right)+{C} \\ $$$$\:\:={x}−\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{arctan}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\mid+{C} \\ $$