Question Number 36965 by MJS last updated on 07/Jun/18
$$\int\frac{{x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}= \\ $$
Commented by rahul 19 last updated on 07/Jun/18
$$\mathrm{is}\:\mathrm{ostradosgki}\:\mathrm{method}\:\mathrm{applicable}\:\mathrm{here}? \\ $$
Commented by MJS last updated on 07/Jun/18
$$\mathrm{I}'\mathrm{ll}\:\mathrm{post}\:\mathrm{a}\:\mathrm{faster}\:\mathrm{method}\:\mathrm{tomorrow} \\ $$
Commented by MJS last updated on 07/Jun/18
$$\mathrm{yes}\:\mathrm{Sir}! \\ $$
Answered by MJS last updated on 07/Jun/18
![Standard Method ∫((x^5 −x^4 +x^3 −1)/((x^2 −x+1)^3 ))dx= =∫((x+1)/(x^2 −x+1))dx−∫(dx/((x^2 −x+1)^2 ))−∫(dx/((x^2 −x+1)^3 ))= ∫((x+1)/(x^2 −x+1))dx=(1/2)∫((2x−1)/(x^2 −x+1))dx+(3/2)∫(dx/(x^2 −x+1))= (1/2)∫((2x−1)/(x^2 −x+1))dx= [t=x^2 −x+1 → dx=(dt/(2x−1))] =(1/2)∫(dt/t)=(1/2)ln t =(1/2)ln(x^2 −x+1) (3/2)∫(dx/(x^2 −x+1))=(3/2)∫(dx/((x−(1/2))^2 +(3/4)))= [t=((√3)/3)(2x−1) → dx=((√3)/2)dt] =(√3)∫(dt/(t^2 +1))=((2(√3))/3)arctan t= =(√3)arctan(((√3)/3)(2x−1)) =(1/2)ln(x^2 −x+1)+(√3)arctan(((√3)/3)(2x−1)) ∫(dx/((x^2 −x+1)^2 ))=16∫(dx/(((2x−1)^2 +3)^2 ))= [t=2x−1 → dx=(dt/2)] =8∫(dt/((t^2 +3)^2 ))= [∫(dt/((at^2 +b)^n ))=(t/(2b(n−1)(at^2 +b)^(n−1) ))+ +((2n−3)/(2b(n−1)))∫(dt/((at^2 +b)^(n−1) ))] =8((t/(6(t^2 +3)))+(1/6)∫(dt/(t^2 +3)))= (1/6)∫(dt/(t^2 +3))= [u=((√3)/3)t → dt=(√3)du] =((√3)/(18))∫(du/(u^2 +1))=((√3)/3)arctan u= =((√3)/(18))arctan(((√3)/3)t) =8((t/(6(t^2 +3)))+((√3)/(18))arctan(((√3)/3)t))= =((2x−1)/(3(x^2 −x+1)))+((4(√3))/9)arctan(((√3)/3)(2x−1)) ∫(dx/((x^2 −x+1)^3 ))=64∫(dx/(((2x−1)^2 +3)^3 ))= [t=2x−1 → dx=(dt/2)] =32∫(dt/((t^2 +3)^3 ))= [∫(dt/((at^2 +b)^n ))=(t/(2b(n−1)(at^2 +b)^(n−1) ))+ +((2n−3)/(2b(n−1)))∫(dt/((at^2 +b)^(n−1) ))] =32((t/(12(t^2 +3)^2 ))+(1/4)∫(dt/((t^2 +3)^2 )))= =32((t/(12(t^2 +3)^2 ))+(1/4)((t/(6(t^2 +3)))+(1/6)∫(dt/(t^2 +3))))= =((8t)/(3(t^2 +3)^2 ))+((4t)/(3(t^2 +3)))+((4(√3))/9)arctan(((√3)/3)t)= =((4t(t^2 +5))/(3(t^2 +3)^2 ))+((4(√3))/9)arctan(((√3)/3)t)= =(((2x−1)(2x^2 −2x+3))/(6(x^2 −x+1)^2 ))+((4(√3))/9)arctan(((√3)/3)(2x−1)) =(1/2)ln(x^2 −x+1)+((√3)/9)arctan(((√3)/3)(2x−1))− −(((2x−1)(4x^2 −4x+5))/(6(x^2 −x+1)^2 ))+C](https://www.tinkutara.com/question/Q36971.png)
$$\mathrm{Standard}\:\mathrm{Method} \\ $$$$\int\frac{{x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}= \\ $$$$=\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}−\int\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }−\int\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}+\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} −{x}+\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}−\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{t}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{3}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:{t}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{3}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\sqrt{\mathrm{3}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{16}\int\frac{{dx}}{\left(\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{2}{x}−\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{8}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)^{{n}} }=\frac{{t}}{\mathrm{2}{b}\left({n}−\mathrm{1}\right)\left({at}^{\mathrm{2}} +{b}\right)^{{n}−\mathrm{1}} }+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{b}\left({n}−\mathrm{1}\right)}\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)^{{n}−\mathrm{1}} }\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{8}\left(\frac{{t}}{\mathrm{6}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{3}}\right)= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{3}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{t}\:\rightarrow\:{dt}=\sqrt{\mathrm{3}}{du}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:{u}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{t}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{8}\left(\frac{{t}}{\mathrm{6}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}+\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{t}\right)\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{64}\int\frac{{dx}}{\left(\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{2}{x}−\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{32}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)^{{n}} }=\frac{{t}}{\mathrm{2}{b}\left({n}−\mathrm{1}\right)\left({at}^{\mathrm{2}} +{b}\right)^{{n}−\mathrm{1}} }+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{b}\left({n}−\mathrm{1}\right)}\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)^{{n}−\mathrm{1}} }\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{32}\left(\frac{{t}}{\mathrm{12}\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{32}\left(\frac{{t}}{\mathrm{12}\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{t}}{\mathrm{6}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{3}}\right)\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{8}{t}}{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{4}{t}}{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{t}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{t}\left({t}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{t}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right)− \\ $$$$\:\:\:\:\:−\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+{C} \\ $$
Answered by MJS last updated on 07/Jun/18
![Ostrogradski′s Method ∫(P/Q)=(P_1 /Q_1 )+∫(P_2 /Q_2 ) P=x^5 −x^4 +x^3 −1 Q=(x^2 −x+1)^3 Q′=3(2x−1)(x^2 −x+1)^2 Q_1 =gcd(Q,Q′)=(x^2 −x+1)^2 Q_2 =(Q/Q_1 )=x^2 −x+1 grade(P_i )<grade(Q_i ) ⇒ P_1 =c_1 x^3 +c_2 x^2 +c_3 x+c_4 P_2 =c_5 x+c_6 differentiate both sides of ∫(P/Q)=(P_1 /Q_1 )+∫(P_2 /Q_2 ) (P/Q)=((P_1 ′Q_1 −P_1 Q_1 ′)/Q_1 ^2 )+(P_2 /Q_2 ) ⇒ (this is the hardest part) ⇒ c_5 =1 −(c_1 +2c_5 −c_6 )=−1 −(c_1 +2c_2 −3c_5 +2c_6 )=1 3c_1 −3c_3 −2c_5 +3c_6 =0 2c_2 +c_3 −4c_4 +c_5 −2c_6 =0 c_3 +2c_4 +c_6 =−1 c_1 =−(4/3); c_2 =3; c_3 =−(7/3); c_4 =(5/6); c_5 =1; c_6 =−(1/3) P_1 =−(4/3)x^3 +2x^2 −(7/3)x+(5/6) P_2 =x−(1/3) ∫((x^5 −x^4 +x^3 −1)/((x^2 −x+1)^3 ))dx =((−(4/3)x^3 +2x^2 −(7/3)x+(5/6))/((x^2 −x+1)^2 ))+∫((x−(1/3))/(x^2 −x+1))dx= =−((8x^3 −12x^2 +14x−5)/(6(x^2 −x+1)^2 ))+∫((3x−1)/(3(x^2 −x+1)))dx= ∫((3x−1)/(3(x^2 −x+1)))dx=(1/3)∫((3x−1)/(x^2 −x+1))dx= (1/3)((3/2)∫((2x−1)/(x^2 −x+1))dx+(1/2)∫(dx/(x^2 −x+1)))= (1/2)∫((2x−1)/(x^2 −x+1))dx+(1/6)∫(dx/(x^2 −x+1))= (1/2)∫((2x−1)/(x^2 −x+1))dx= [t=x^2 −x+1 → dx=(dt/(2x−1))] =(1/2)∫(dt/t)=(1/2)ln t=(1/2)ln(x^2 −x+1) (1/6)∫(dx/(x^2 −x+1))=(1/6)∫(dx/((x−(1/2))^2 +(3/4)))= [t=((√3)/3)(2x−1) → dx=((√3)/2)dt] ((√3)/9)∫(dt/(t^2 +1))=((√3)/9)arctan t= =((√3)/9)arctan(((√3)/3)(2x−1)) =(1/2)ln(x^2 −x+1)+((√3)/9)arctan(((√3)/3)(2x−1)) =(1/2)ln(x^2 −x+1)+((√3)/9)arctan(((√3)/3)(2x−1))− −((8x^3 −12x^2 +14x−5)/(6(x^2 −x+1)^2 ))+C](https://www.tinkutara.com/question/Q37013.png)
$$\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method} \\ $$$$\int\frac{{P}}{{Q}}=\frac{{P}_{\mathrm{1}} }{{Q}_{\mathrm{1}} }+\int\frac{{P}_{\mathrm{2}} }{{Q}_{\mathrm{2}} } \\ $$$${P}={x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{1} \\ $$$${Q}=\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${Q}'=\mathrm{3}\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${Q}_{\mathrm{1}} =\mathrm{gcd}\left({Q},{Q}'\right)=\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${Q}_{\mathrm{2}} =\frac{{Q}}{{Q}_{\mathrm{1}} }={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$$\mathrm{grade}\left({P}_{{i}} \right)<\mathrm{grade}\left({Q}_{{i}} \right)\:\Rightarrow \\ $$$${P}_{\mathrm{1}} ={c}_{\mathrm{1}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{3}} {x}+{c}_{\mathrm{4}} \\ $$$${P}_{\mathrm{2}} ={c}_{\mathrm{5}} {x}+{c}_{\mathrm{6}} \\ $$$$\mathrm{differentiate}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{of}\:\int\frac{{P}}{{Q}}=\frac{{P}_{\mathrm{1}} }{{Q}_{\mathrm{1}} }+\int\frac{{P}_{\mathrm{2}} }{{Q}_{\mathrm{2}} } \\ $$$$\frac{{P}}{{Q}}=\frac{{P}_{\mathrm{1}} '{Q}_{\mathrm{1}} −{P}_{\mathrm{1}} {Q}_{\mathrm{1}} '}{{Q}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{{P}_{\mathrm{2}} }{{Q}_{\mathrm{2}} }\:\Rightarrow \\ $$$$\left(\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{hardest}\:\mathrm{part}\right) \\ $$$$\Rightarrow \\ $$$${c}_{\mathrm{5}} =\mathrm{1} \\ $$$$−\left({c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{5}} −{c}_{\mathrm{6}} \right)=−\mathrm{1} \\ $$$$−\left({c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{2}} −\mathrm{3}{c}_{\mathrm{5}} +\mathrm{2}{c}_{\mathrm{6}} \right)=\mathrm{1} \\ $$$$\mathrm{3}{c}_{\mathrm{1}} −\mathrm{3}{c}_{\mathrm{3}} −\mathrm{2}{c}_{\mathrm{5}} +\mathrm{3}{c}_{\mathrm{6}} =\mathrm{0} \\ $$$$\mathrm{2}{c}_{\mathrm{2}} +{c}_{\mathrm{3}} −\mathrm{4}{c}_{\mathrm{4}} +{c}_{\mathrm{5}} −\mathrm{2}{c}_{\mathrm{6}} =\mathrm{0} \\ $$$${c}_{\mathrm{3}} +\mathrm{2}{c}_{\mathrm{4}} +{c}_{\mathrm{6}} =−\mathrm{1} \\ $$$$ \\ $$$${c}_{\mathrm{1}} =−\frac{\mathrm{4}}{\mathrm{3}};\:{c}_{\mathrm{2}} =\mathrm{3};\:{c}_{\mathrm{3}} =−\frac{\mathrm{7}}{\mathrm{3}};\:{c}_{\mathrm{4}} =\frac{\mathrm{5}}{\mathrm{6}};\:{c}_{\mathrm{5}} =\mathrm{1};\:{c}_{\mathrm{6}} =−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${P}_{\mathrm{1}} =−\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{3}}{x}+\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${P}_{\mathrm{2}} ={x}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$\int\frac{{x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$$$=\frac{−\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{3}}{x}+\frac{\mathrm{5}}{\mathrm{6}}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+\int\frac{{x}−\frac{\mathrm{1}}{\mathrm{3}}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=−\frac{\mathrm{8}{x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{2}} +\mathrm{14}{x}−\mathrm{5}}{\mathrm{6}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+\int\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} −{x}+\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}−\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{t}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:{t}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)\right)− \\ $$$$\:\:\:\:\:−\frac{\mathrm{8}{x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{2}} +\mathrm{14}{x}−\mathrm{5}}{\mathrm{6}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+{C} \\ $$