Evaluate ∫1x2−x+1dx

∫1x2−x+1dx=∫1(x−1/2)2+3/4dx=∫1(x−1/2)2+(3/2)2dx=13/2tan−1⁡x−1/23/2+C=23tan−1⁡2x−13+C

Evaluate ∫1x2−x+1dx

∫1x2−x+1dx=∫1(x−1/2)2+3/4dx=∫1(x−1/2)2+(3/2)2dx=13/2tan−1⁡x−1/23/2+C=23tan−1⁡2x−13+C

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

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

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

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} \int \frac{1}{x^{2} - x + 1} d x \\ = \int \frac{1}{(x - 1 / 2)^{2} + 3 / 4} d x \\ = \int \frac{1}{(x - 1 / 2)^{2} + (\sqrt{3} / 2)^{2}} d x \\ = \frac{1}{\sqrt{3} / 2} \tan^{- 1} (\frac{x - 1 / 2}{\sqrt{3} / 2}) + C \\ = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C \end{array}$