The Ultimate Test™

$\begin{aligned} I & = \int_{-1}^1 \lim_{y \to \infty} \left(\tan^{-1} \left(\sum_{z=1}^y x^{2x}\ln (y-z+1)\right) + \color{#3D99F6} \tan^{-1} \left(\frac 1{\sum_{z=1}^y x^{2x}\ln (y-z+1)} \right) \right) dx \\ & = \int_{-1}^1 \lim_{y \to \infty} \left(\tan^{-1} \left(\sum_{z=1}^y x^{2x}\ln (y-z+1)\right) + \color{#3D99F6} \frac \pi 2 - \tan^{-1} \left(\sum_{z=1}^y x^{2x}\ln (y-z+1)\right) \right) dx \\ & = \int_{-1}^1 \frac \pi 2 \ dx = \frac \pi 2 x \ \bigg|_{-1}^1 = \pi \approx \boxed{3.142} \end{aligned}$

From easy trigonometry in a right angled triangle you can see that

$arctan(a)+arctan\left(\frac{1}{a}\right)=\frac{\pi}{2}$

The y limit is seen to be irrelevant and the integral then simplifies to

$\int_{-1}^{1}\frac{\pi}{2}dx=\boxed{\pi}$