Question Number 65988 by mmkkmm000m last updated on 07/Aug/19
![∫dx/x^2 −x+1](https://www.tinkutara.com/question/Q65988.png)
$$\int{dx}/{x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 07/Aug/19
![let A =∫ (dx/(x^2 −x+1)) ⇒A =∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) (4/3) ∫ (1/(1+t^2 )) ((√3)/2)dt =(2/( (√3)))arctan(((2x−1)/( (√3))))+c .](https://www.tinkutara.com/question/Q65997.png)
$${let}\:{A}\:=\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\Rightarrow{A}\:=\int\:\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c}\:. \\ $$
Answered by meme last updated on 07/Aug/19
![= −∫−(1/x^2 )dx−x+1 = −(1/x) −x+1 = −((1−x+x^2 )/x)](https://www.tinkutara.com/question/Q65993.png)
$$\:\:\:\:\:=\:\:\:−\int−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx}−{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:=\:−\frac{\mathrm{1}}{{x}}\:−{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:=\:−\frac{\mathrm{1}−{x}+{x}^{\mathrm{2}} }{{x}} \\ $$$$\:\:\:\:\:\: \\ $$
Answered by meme last updated on 07/Aug/19
![b^2 −4ac=1−4=−3<0 ∫(dx/(1−x+x^2 ))= (2/( (√3)))arctan(((2x−1)/( (√3))))+c](https://www.tinkutara.com/question/Q66000.png)
$$\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{1}−\mathrm{4}=−\mathrm{3}<\mathrm{0} \\ $$$$\:\:\:\:\:\int\frac{{dx}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c} \\ $$
Answered by AnjanDey last updated on 07/Aug/19
![=∫(dx/(x^2 −2∙x∙(1/2)+((1/2))^2 +1−((1/2))^2 )) =∫(dx/((x−(1/2))^2 +(((√3)/2))^2 )) =(2/( (√3)))tan^(−1) (((x−(1/2))/((√3)/2)))+C =(2/( (√3)))tan^(−1) (((2x−1)/( (√3))))+C](https://www.tinkutara.com/question/Q66020.png)
$$=\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}\centerdot{x}\centerdot\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)+{C} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$