I'm trying to solve \(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{1}}}{{{\left({4}+{x}^{{2}}\right)}\sqrt{{{4}+{x}^{{2}}}}}}}{\left.{d}{x}\right.}\)

Therefore, after substitution, the integral becomes
${\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{\sqrt{1+{\tan }^{2}t}},dt}$
To proceed further, you need to use the following cousin of the Pythagorean theorem:
${\displaystyle 1+{\tan }^{2}t={\sec }^{2}t}$
This identity is extremely important and useful in practice -- not the least in manipulating integrals like this. One should be reminded of this identity whenever one comes across an expression like ${\displaystyle \sqrt{1+{\tan }^{2}t}}$ or ${\displaystyle \sqrt{1+x^2}}$. By the way, the proof of this identity is based on the standard Pythagorean theorem:
${\displaystyle 1+{\tan }^{2}t=1+\frac{{\sin }^{2}t}{{\cos }^{2}t}=\frac{{\cos }^{2}t+{\sin }^{2}t}{{\cos }^{2}t}=\frac{1}{{\cos }^{2}t}={\sec }^{2}t}$
From this, it follows that
${\displaystyle \frac{1}{\sqrt{1+{\tan }^{2}t}}=\left|\cos t\right|}$
Thus the integral becomes
${\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left|\cos t\right|,dt}$