Elegant $\arctan$

I have come up with an elementary approach to solve this integral (no feynman’s, infinite series etc.)

I $= \int_0^{\infty} \arctan(\frac{1}{x^2})dx$

Let $u = \frac{1}{x}$ and then setting $u = x$ ,

I $= \int_0^{\infty} \frac{arctan(x^2)}{x^2}dx$

Taking $arctan(x^2)$ as the first function and $\frac{1}{x^2}$ as the second one in integration by parts,

I $= [\arctan(x^2) \int \frac{1}{x^2}dx + \int \frac{2}{1+x^4}dx]_0^{\infty}$

First term approaches 0 as $x \to 0$ or $x \to \infty$ . The second term can be solved through partial fraction decomposition or trig substitution.

The answer comes out to be $\frac{\sqrt{2}\pi}{2}$

Similar solution with @Aaghaz Mahajan 's

$\begin{aligned} I(a) & = \int_0^\infty \tan^{-1} \left(\frac a{x^2} \right) dx \\ \frac {\partial I(a)}{\partial a} & = \int_0^\infty \frac {x^2}{x^4 + a^2} dx \\ & = \frac 12 \int_0^\infty \frac 1{x^2 + {\color{#3D99F6}i} a} + \frac 1{x^2-{\color{#3D99F6}i} a} dx & \small \color{#3D99F6} \text{where }i = \sqrt{-1} \text{ denotes the imaginary unit.} \\ & = \frac 12 \left[\frac 1{\sqrt{ia}} \tan^{-1} \left(\frac x{\sqrt{ia}} \right) + \frac 1{\sqrt{-ia}} \tan^{-1} \left(\frac x{\sqrt{-ia}} \right) \right]_0^\infty \\ & = \frac \pi 4 \left[\frac 1{\sqrt{ia}} + \frac 1{\sqrt{-ia}} \right] = \frac \pi {2\sqrt{2a}} & \small \color{#3D99F6} \text{Integrate both sides w.r.t. }a \\ \implies I(a) & = \frac \pi {\sqrt 2}\sqrt a + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ I(a) & = \frac \pi {\sqrt 2}\sqrt a & \small \color{#3D99F6} \text{Since }I(0) = 0 \implies C = 0 \\ \implies I(1) & = \int_0^\infty \tan^{-1} \left(\frac 1{x^2} \right) dx = \frac \pi {\sqrt 2} = \frac {\sqrt 2\pi}2 \end{aligned}$

Therefore, $\min (A+B+C) = 1+2+2 = \boxed 5$ .

Let

$I\left(a\right)=\int_0^{\infty}\arctan\left(\frac{a}{x^2}\right)dx$

So,

$I'\left(a\right)=\int_0^{\infty}\frac{dx}{\left(x^2+\frac{a^2}{x^2}\right)}$

Putting $\displaystyle x=\sqrt{a}t$ we get

$I'\left(a\right)=\frac{1}{\sqrt{a}}\int_0^{\infty}\frac{dt}{\left(t^2+\frac{1}{t^2}\right)}$

And so,

$I'\left(a\right)=\frac{\pi}{2\sqrt{a}\sqrt{2}}$

Now, integrating and using $\displaystyle I\left(0\right)=0$ we get

$I\left(a\right)=\pi\sqrt{\frac{a}{2}}$

Thus, making the answer

$I\left(1\right)=\frac{\pi}{\sqrt{2}}=\frac{\sqrt{2}\pi^1}{2}$