Correct Answer - Option 3 :
\(\frac{\pi}{\sqrt{2}}\)
Concept:
Some standard integration formulae:
\(\rm \Rightarrow \int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}+c\)
\(\rm\Rightarrow \int \frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}log\left | \frac{x-a}{x+a} \right |+c\)
\(\rm \Rightarrow \int \frac{dx}{a^{2}-x^{2}}=\frac{1}{2a}log\left | \frac{a+x}{a-x} \right |+c\)
Calculation:
\(\rm\int_{-\infty }^{\infty }\frac{dx}{2+x^{2}}\\=\int_{-\infty }^{\infty}\frac{dx}{\left ( \sqrt{2} \right )^{2}+x^{2}}\)
\(\rm=\frac{1}{\sqrt{2}}tan^{-1}\frac{x}{\sqrt{2}}\left.\begin{matrix} \\ \end{matrix}\right|_{-\infty }^{\infty }\)
\(\rm=\frac{1}{\sqrt{2}}\left ( tan ^{-1}\left ( \frac{\infty }{\sqrt{2}} \right )-tan^{-1}\left ( \frac{-\infty }{\sqrt{2}} \right )\right )\)
\(\rm=\frac{1}{\sqrt{2}}\left ( \frac{\pi}{2} -\left ( \frac{-\pi}{2} \right )\right )\)
\(=\frac{\pi}{\sqrt{2}}\)
∴ The correct option is 3
