For $a\ge 2,$ if the value of the definite integral $\underset{0}{\overset{\infty }{\int }}\frac{dx}{a^2+{\left(x-\left(\frac{1}{x}\right)\right)}^2}$ equals $\frac{\pi }{10}$ ,then the value of a is

$$\begin{gathered} I=\underset{0}{\overset{\infty }{\int }}\frac{dx}{x^2+\frac{1}{x^2}+(a^2-2)}=\underset{0}{\overset{\infty }{\int }}\frac{x^{2}dx}{x^4+(a^2-2)x^2+1},Put{a}^2-2=k \\ =\underset{0}{\overset{\infty }{\int }}\frac{x^{2}dx}{x^4+k{x}^2+1}=\frac{1}{2}\underset{0}{\overset{\infty }{\int }}\frac{(x^2+1)+(x^2-1)}{x^4+k{x}^2+1}dx \\ =\frac{1}{2}\underset{I_1}{\underbrace{\underset{0}{\overset{\infty }{\int }}\frac{1+\left(\frac{1}{x^2}\right)}{x^2+\left(\frac{1}{x^2}\right)+k}}}dx+\frac{1}{2}\underset{I_2}{\underbrace{\underset{0}{\overset{\infty }{\int }}\frac{1-\left(\frac{1}{x^2}\right)}{x^2+\left(\frac{1}{x^2}\right)+k}}}dx \end{gathered}$$

Now proceed,  ${\text{I}}_{\text{1}}\text{=}\frac{\text{π}}{\text{2a}}{\text{\hspace{0.17em}and\hspace{0.17em}I}}_{\text{2}}\text{=0,}$

$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I=}\frac{\pi }{2a}=\frac{\pi }{10}$

$\Rightarrow a=5$