∫_0^1dx/(x^2+x+1)

$I=\int_{0}^{1}\frac{dx}{(x^2+x+\frac{1}{4})-\frac{1}{4}+1}$

$=\int_{0}^{1}\frac{dx}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$

$=\frac{2}{\sqrt{3}}\begin{bmatrix}tan^{-1}\frac{(x+\frac{1}{2})}{\frac{\sqrt{3}}{2}}\end{bmatrix}_0^1$

$=\frac{2}{\sqrt{3}}\begin{bmatrix}tan^{-1}(\frac{3}{2}×\frac{2}{\sqrt{3}})-tan^{-1}(\frac{1}{2}×\frac{2}{\sqrt{3}})\end{bmatrix}$

$=\frac{2}{\sqrt{3}}\begin{bmatrix}tan^{-1}(\frac{3}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}})-tan^{-1}(\frac{1}{\sqrt{3}})\end{bmatrix}$

$=\frac{2}{\sqrt{3}}\begin{bmatrix}tan^{-1}(\frac{3\sqrt{3}}{3})-tan^{-1}(\frac{1}{\sqrt{3}})\end{bmatrix}$

$=\frac{2}{\sqrt{3}}\begin{bmatrix}tan^{-1}(\sqrt{3})-tan^{-1}(\frac{1}{\sqrt{3}})\end{bmatrix}$

$=\frac{2}{\sqrt{3}}\begin{bmatrix}\frac{\pi}{3}-\frac{\pi}{6}\end{bmatrix}$ (∵ $tan\frac{\pi}{3}=\sqrt{3}, tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$)

$⇒ \frac{2}{\sqrt{3}}×\frac{2\pi - \pi}{6}$

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

Option 1 is correct.