Question

$\underset{0}{\overset{1}{\int }}\frac{1}{x^2+x+1}dx$ is equal to

• A. $\frac{\pi }{\sqrt{3}}$
• B. $\frac{\pi }{2\sqrt{3}}$
• C. $\frac{\pi }{3\sqrt{3}}$
• D. $\frac{\pi }{4\sqrt{3}}$

Answer 1

The correct option is C $\frac{\pi }{3\sqrt{3}}$

Let $I=\underset{0}{\overset{1}{\int }}\frac{1}{x^2+x+1}dx=\underset{0}{\overset{1}{\int }}\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}dx\{\because x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}\}={\left[\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right]}_0^1\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}=\frac{2}{\sqrt{3}}({\tan }^{-1}\sqrt{3}-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right))=\frac{2}{\sqrt{3}}\left(\frac{\pi }{6}\right)=\frac{\pi }{3\sqrt{3}}$