First slide

Introduction to integration

Question

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

Easy

A

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

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

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

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

Solution

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

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