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# (a) By completing the square, show that $\int^{1/2}_0 \frac {dx}{x^2 - x + 1} = \frac {\pi}{3 \sqrt 3}$(b) By factoring $x^3 + 1$ as a sum of cubes, rewrite the integral in part (a). Then express $1/(x^3 + 1)$ as the sum of a power series and use it to prove the following formula for $\pi:$ $\pi = \frac {3 \sqrt 3 }{4} \sum_{n = 0}^{\infty} \frac {(-1)^n}{8^n} \left( \frac {2}{3n + 1} + \frac {1}{3n + 2} \right)$

## a. $=\frac{2}{\sqrt{3}} \times \frac{\pi}{6}=\frac{\pi}{3 \sqrt{3}}$b. $$=\frac{3 \sqrt{3}}{4} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{8^{n}}\left(\frac{2}{3 n+1}+\frac{1}{3 n+2}\right)$$

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