dx-x-2-x-1-

Question Number 65988 by mmkkmm000m last updated on 07/Aug/19

\int d x / x^{2} - x + 1

Commented by mathmax by abdo last updated on 07/Aug/19

l e t A = \int \frac{d x}{x^{2} - x + 1} \Rightarrow A = \int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}

=_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} t} \frac{4}{3} \int \frac{1}{1 + t^{2}} \frac{\sqrt{3}}{2} d t = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + c .

Answered by meme last updated on 07/Aug/19

= - \int - \frac{1}{x^{2}} d x - x + 1

= - \frac{1}{x} - x + 1

= - \frac{1 - x + x^{2}}{x}

Answered by meme last updated on 07/Aug/19

b^{2} - 4 a c = 1 - 4 = - 3 < 0

\int \frac{d x}{1 - x + x^{2}} = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + c

Answered by AnjanDey last updated on 07/Aug/19

= \int \frac{d x}{x^{2} - 2 \cdot x \cdot \frac{1}{2} + {(\frac{1}{2})}^{2} + 1 - {(\frac{1}{2})}^{2}}

= \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}

= \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C

= \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C

Facebook Tweet Pin