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

$$\int dx/x^2-x+1$$

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

$$letA=\int \frac{dx}{x^2-x+1}\Rightarrow A=\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}$$$${=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}t}\frac{4}{3}\int \frac{1}{1+t^2}\frac{\sqrt{3}}{2}dt=\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1}{\sqrt{3}}\right)+c.$$

Answered by meme last updated on 07/Aug/19

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

Answered by meme last updated on 07/Aug/19

$$b^2-4ac=1-4=-3<0$$$$\int \frac{dx}{1-x+x^2}=\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1}{\sqrt{3}}\right)+c$$

Answered by AnjanDey last updated on 07/Aug/19

$$=\int \frac{dx}{x^2-2\cdot x\cdot \frac{1}{2}+{\left(\frac{1}{2}\right)}^2+1-{\left(\frac{1}{2}\right)}^2}$$$$=\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$$$$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$$

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