Question

${\int }_0^1\frac{dx}{x^2+x+1}$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{3\sqrt{3}}$
• C. $\frac{\pi }{\sqrt{3}}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (B)

$\frac{\pi }{3\sqrt{3}}$

${\int }_0^1\frac{dx}{x^2+x+1}$

$={\int }_0^1\frac{dx}{x^2+x+\frac{1}{4}-\frac{1}{4}+1}$

$={\int }_0^1\frac{dx}{{(x+\frac{1}{2})}^2-\frac{1}{4}+1}$

$={\int }_0^1\frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$

$={\int }_0^1\frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$={\left[\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]}_0^1$

$={\left[\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]}_0^1$

$=\frac{2}{\sqrt{3}}[{\tan }^{-1}\sqrt{3}-{\tan }^{-1}\frac{1}{\sqrt{3}}]$

$=\frac{2}{\sqrt{3}}[\frac{\pi }{3}-\frac{\pi }{6}]$

$=\frac{2}{\sqrt{3}}\left(\frac{\pi }{6}\right)$

$=\frac{\pi }{3\sqrt{3}}$

Hence the correct answer is $\left(B\right)\frac{\pi }{3\sqrt{3}}$.