This cannot be real!

$\begin{aligned} I & = \int_0^\infty \frac 1{1+\tan^2\left(2\tan^{-1}\left(\frac {\sqrt{1+x^2}-1}x\right)\right)} dx \\ & = \int_0^\infty \left(1+\tan^2 \left(\tan^{-1}\left(\frac {2\left(\frac {\sqrt{1+x^2}-1}x\right)}{1-\left( \frac {\sqrt{1+x^2}-1}x \right)^2}\right)\right)\right)^{-1} dx \\ & = \int_0^\infty \left(1+\tan^2 \left(\tan^{-1} x\right)\right)^{-1} dx \\ & = \int_0^\infty \frac 1{1+x^2} dx \\ & = \tan^{-1} x \ \bigg|_0^\infty = \frac \pi 2 \approx \boxed{1.571} \end{aligned}$

Let $x=\tan α$ . Then $dx=\sec^2 αdα$ and the integrand is $\cos^2 α$ . The limits of integration become $0$ and $\dfrac{π}{2}$ . So the value of the integral is $\boxed {\dfrac{π}{2}}$

Firstly, note that $\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta} \implies 2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ .

Knowing this and $1+\tan^2x=\sec^2x$ , $\frac{1}{1+\tan^2\left(2\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\right)}=\frac{1}{\sec^2\left(\cos^{-1}\left(\frac{1-\left(\frac{\sqrt{1+x^2}-1}{x}\right)^2}{1+\left(\frac{\sqrt{1+x^2}-1}{x}\right)^2}\right)\right)}$ $=\cos^2\left(\cos^{-1}\left(\frac{x^2}{x^2}\cdot\frac{1-\frac{x^2+2-2\sqrt{1+x^2}}{x^2}}{1+\frac{x^2+2+2\sqrt{1+x^2}}{x^2}}\right)\right)$ $=\left(\frac{1+\sqrt{1+x^2}}{x^2+1+\sqrt{1+x^2}}\right)^2$ $=\frac{x^2+2\sqrt{x^2+1}+2}{x^4+3x^2+2+2x^2\sqrt{x^2+1}+2\sqrt{x^2+1}}$

We can use trigonometric substitution to integrate this monstrosity: $x=\tan\theta;~dx=\sec^2\theta d\theta$ $\arctan0=0;~\displaystyle\lim_{x\rightarrow \infty}\arctan x=\frac{\pi}{2}$ $\ \displaystyle\int_{0}^{\infty}\frac{x^2+2\sqrt{x^2+1}+2}{x^4+3x^2+2+2x^2\sqrt{x^2+1}+2\sqrt{x^2+1}}dx=\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\tan^2\theta+2\sec\theta+2}{\tan^4\theta+3\tan^2\theta+2+2\tan^2\theta\sec\theta+2\sec\theta}\sec^2\theta~d\theta$ $=\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sec^2\theta+2\sec\theta+1}{(\tan^2+1)(\tan^2+2)+2\sec\theta(\tan^2\theta+1)}\sec^2\theta~d\theta$ $=\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{(\sec\theta+1)^2}{\sec^2\theta(\sec^2\theta+1)+\sec^3\theta}\sec^2\theta~d\theta$ $=\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{(\sec\theta+1)^2}{\sec^2\theta+1+2\sec\theta}d\theta$ $=\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{(\sec\theta+1)^2}{(\sec\theta+1)^2}d\theta=\displaystyle\int_{0}^{\frac{\pi}{2}}d\theta=\frac{\pi}{2}-0=\boxed{\frac{\pi}{2}}$ ${\beta_{\lceil \mid \rceil}}$

Consider the expression:

$A = \frac{\sqrt{1+x^2}-1}{x}$

Let $x = \tan{t}$ . This leads to:

$A = \frac{\sec{t}-1}{\tan{t}} = \frac{1-\cos{t}}{\sin{t}} = \frac{2\sin^2\left(\frac{t}{2}\right)}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} = \tan\left(\frac{t}{2}\right) \implies \tan(2\arctan{A}) = \tan{t} = x$

Using this, the integrand simplifies to:

$I = \int_{0}^{\infty}\frac{1}{1+x^2}dx$

Recall that $x = \tan{t} \implies dx = \sec^2{t}dt$ and the integral transforms to:

$I = \int_{0}^{\frac{\pi}{2}}dt \implies \boxed{I = \frac{\pi}{2}}$

$\tan\left(2\arctan\left(\displaystyle\frac{\sqrt{1+x^{2}}-1}{x}\right)\right)=x$

$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{2}} dx =\frac{\pi}{2}$

We can derive this by doing integration by parts on $\displaystyle \int \frac{1}{1+x^{2}} dx$

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

By changing variables from $x=\tan\theta$ and finding the anti-derivative and converting back to original variable and equating this integral with $\arctan x$ we will get the following equation: $\sin\left(2\theta\right)=\frac{2x}{1+x^{2}}$

where $\theta =\arctan x$ from the variable change

$\displaystyle \arcsin\left(\frac{2x}{1+x^{2}}\right)=2\arctan x$

But we also know that $\displaystyle \arcsin x =\arctan\left(\frac{x}{\sqrt{1-x^{2}}}\right)$

Using this two equations and rearranging variable we can arrive at the following equation $\tan\left(2\arctan\left(\displaystyle\frac{\sqrt{1+x^{2}}-1}{x}\right)\right)=x$ .