How to compute $\int_{0}^{1}\left (\frac{\arctan x}{1+(x+\frac{1}{x})\arctan x}\right )^2dx$

Here is a way to arrive at the answer, and it is by no means obvious.

In finding the indefinite integral the reverse quotient rule will be used. Recall that if $u$ and $v$ are differentiable functions, from the quotient rule $$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$ it is immediate that $$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$

Now the indefinite integral $$I = \int \left (\frac{\arctan (x)}{1 + \left (x + \frac{1}{x} \right ) \arctan (x)} \right )^2 \, dx,$$ can be rewritten as $$I = \int \frac{x^2 \arctan^2 (x)}{\left (x + (x^2 + 1) \arctan (x) \right )^2} \, dx.$$

Let $v = x + (x^2 + 1) \arctan (x)$. Then $v' = 2 + 2x \arctan (x)$. Now for the hard bit. We need to find a function $u(x)$ such that $$u' v - v' u = u'[x + (x^2 + 1) \arctan (x)] - u [2 - 2x \arctan (x)] = x^2 \arctan^2 (x).$$ If $$u = \frac{-x^2 + (x^2 + 1) \arctan^2 (x)}{2},$$ then $$u' = x + x \arctan^2 (x) + \arctan (x),$$ and we find, miraculously, that $$u' v - v' u = x^2 \arctan^2 (x).$$

Our indefinite integral can now be readily found as it can be rewritten in the form given by (1). The result is: $$I = \int \left (\frac{-x^2 + (1 + x^2) \arctan^2 (x)}{2[x + (x^2 + 1) \arctan (x)]} \right )' \, dx = \frac{-x^2 + (1 + x^2) \arctan^2 (x)}{2[x + (x^2 + 1) \arctan (x)]} + C.$$

So for the definite integral on the interval $x \in [0,1]$ we have $$\int_0^1 \left (\frac{\arctan (x)}{1 + \left (x + \frac{1}{x} \right ) \arctan (x)} \right )^2 \, dx = \frac{\pi^2 - 8}{8(2 + \pi)}.$$

In fact, there exists a closed-form expression for the indefinite integral (credit Mathematica, not me, for this): $${-x^2+(1+x^2)(\arctan(x))^2\over 2(x+(1+x^2)\arctan(x))}\,,$$ as can be verified by differentiation. Integral over $x\in [0,1]$ is equal to ${\pi^2-8\over 8(2+\pi)}\approx0.0454529$.

This is about as far as I got. Let me know if anyone can finish this analytically:

$$ \left( \frac{\arctan x}{1 + (x + \frac{1}{x}) \arctan x} \right)^2 = \\ \left( \frac{\frac{x}{1 + x^2} \cdot \arctan x}{x \cdot \frac{1}{1 + x^2} + \arctan x} \right)^2 = \\ \left( \frac{x \arctan x \cdot (\arctan x)'}{(x \arctan x)'} \right)^2 = \\ \left( \frac{(\arctan x)'}{[\ln(x \arctan x)]'} \right)^2 $$