Question

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

• 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

The correct option is D

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

Explanation for correct option:

Let $\mathrm{I}={\int }_0^1\frac{1}{\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}d\mathrm{x}$

Modifying the denominator to make it a perfect square, so that we can apply a formula $x^2+a^2$

Hence adding and subtracting $\frac{1}{4}$in the denominator.

$$\begin{gathered} \mathrm{I}={\int }_0^1\frac{1}{1+\left({\mathrm{x}}^2+\mathrm{x}+{\displaystyle \frac{1}{4}}\right)-{\displaystyle \frac{1}{4}}}d\mathrm{x} \\ \mathrm{I}={\int }_0^1\frac{1}{{\left(\mathrm{x}+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}d\mathrm{x} \end{gathered}$$

Here, $a=\frac{\sqrt{3}}{2}$.

Now we know that, $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$

Therefore,

$\begin{array}{rcl}\mathrm{I}& =& \frac{1}{\frac{\sqrt{3}}{2}}\mathrm{t}{{\mathrm{an}}^{-1}\left(\frac{\mathrm{x}+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)}_0^1\\ & =& \frac{2}{\sqrt{3}}\left[{\tan }^{-1}\left(\frac{3}{2}\times \frac{2}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{1}{2}\times \frac{2}{\sqrt{3}}\right)\right]\\ & =& \frac{2}{\sqrt{3}}\left[{\tan }^{-1}\left(\sqrt{3}\right)-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\right]\\ & =& \frac{2}{\sqrt{3}}\left[\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\right]\\ & =& \frac{2}{\sqrt{3}}\left[\frac{3\mathrm{\pi }}{6}\right]\\ \mathrm{I}& =& \frac{\mathrm{\pi }}{3\sqrt{3}}\end{array}$

Hence, option (D) is the correct answer.