A calculus problem by Akshat Sharda

$\begin{aligned} Q & = \; \Big[\tfrac12 \tan^{-1}x^2\Big]_0^\infty \; =\; \tfrac14\pi \\ P - R & = \; \int_0^\infty \frac{x^2-1}{x^4+1}\,dx \; = \; \int_0^\infty \frac{1 - x^{-2}}{(x + x^{-1})^2-2}\,dx \; = \; \int_0^\infty \frac{d}{dx}\left(\frac{1}{2\sqrt{2}}\ln\left(\frac{x + x^{-1} - \sqrt{2}}{x + x^{-1} + \sqrt{2}}\right)\right)\,dx \\ & = \; \Big[\frac{1}{2\sqrt{2}}\ln\left(\frac{x + x^{-1} - \sqrt{2}}{x + x^{-1} - \sqrt{2}}\right)\Big]_0^\infty \; = \; 0 \\ P + Q & = \; \int_0^\infty \frac{x^2+1}{x^4+1}\,dx \; = \; \int_0^\infty \frac{1 + x^{-2}}{(x - x^{-1})^2+2}\,dx \; = \; \int_0^\infty \frac{d}{dx}\left(\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x - x^{-1}}{\sqrt{2}}\right)\right)\,dx \\ & = \; \Big[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x - x^{-1}}{\sqrt{2}}\right)\Big]_0^\infty \; = \; \frac{\pi}{\sqrt{2}} \end{aligned}$ so that $Q = \tfrac14\pi$ , $P=R$ and $P+R - Q\sqrt{2} = \frac{\pi}{2\sqrt{2}}$ .