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Question Number 77552 by lémùst last updated on 07/Jan/20

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

Answered by MJS last updated on 07/Jan/20

$$\int \frac{x-2}{x^2-x+1}dx=\frac{1}{2}\int \frac{2x-4}{x^2-x+1}dx=$$$$=\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx-\frac{3}{2}\int \frac{dx}{x^2-x+1}=$$$$$$$$\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx=\frac{1}{2}\ln (x^2-x+1)$$$$-\frac{3}{2}\int \frac{dx}{x^2-x+1}=-\frac{3}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}=$$$$[t=\frac{\sqrt{3}}{3}(2x-1)\to dx=\frac{\sqrt{3}}{2}dt]$$$$=-\sqrt{3}\int \frac{dt}{t^2+1}=-\sqrt{3}\arctan t=-\sqrt{3}\arctan \frac{\sqrt{3}(2x-1)}{3}$$$$$$$$=\frac{1}{2}\ln (x^2-x+1)-\sqrt{3}\arctan \frac{\sqrt{3}(2x-1)}{3}+C$$

Commented by lémùst last updated on 07/Jan/20

$$\mathrm{thank}\mathrm{you}!$$

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