Question

${\int }_0^1\frac{dx}{1+x+x^2}$ is equal to.

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

Answer 1

Answer: (D)

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

$I={\int }_0^1\frac{dx}{1+x+x^2}$
$={\int }_0^1\frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$ (making squaring the term)
Using, $I=\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$
$\therefore I={\int }_0^1\frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=\frac{2}{\sqrt{3}}{\left[{\tan }^{-1}\frac{(x+\frac{1}{2})}{\left(\frac{\sqrt{3}}{2}\right)}\right]}_0^1$
$=\frac{2}{\sqrt{3}}[\frac{\pi }{3}-\frac{\pi }{6}]$
$=\frac{\pi }{3\sqrt{3}}$ Option D is correct