Over 3600 Years From Ahmes To ... ?

To begin with, note that $\int_0^1 \frac{1}{(1+x^2)\sqrt{2+x^2}}\,dx \; = \; \left[ \tan^{-1}\left(\frac{x}{\sqrt{2+x^2}}\right) \right]_0^1 \; = \; \tan^{-1}\tfrac{1}{\sqrt{3}} \; = \; \tfrac16\pi$ If we define $f(y) \; = \; \int_0^1 \frac{\tan^{-1}\big(y\sqrt{2+x^2}\big)}{(1+x^2)\sqrt{2+x^2}}\,dx$ then we see that $\lim_{y\to\infty}f(y) \; =\; \tfrac12\pi\int_0^1 \frac{1}{(1+x^2)\sqrt{2+x^2}}\,dx \; = \; \tfrac{1}{12}\pi^2 \;,$ while, differentiating inside the integral sign, $\begin{array}{rcl} \displaystyle f'(y) & = & \displaystyle \int_0^1 \frac{1}{(1+x^2)(1 + 2y^2 + y^2x^2)}\,dx \\ & = & \displaystyle \frac{1}{1+y^2}\int_0^1\left\{ \frac{1}{1+x^2} - \frac{y^2}{1 + 2y^2 + y^2x^2}\right\}\,dx \\ & = & \displaystyle \frac{1}{1+y^2}\left[ \tan^{-1}x - \frac{y}{\sqrt{1+2y^2}}\tan^{-1}\left(\frac{yx}{\sqrt{1+2y^2}}\right)\right]_0^1 \\ & = & \displaystyle \tfrac14\pi \frac{1}{1+y^2} - \frac{y}{(1+y^2)\sqrt{1+2y^2}}\tan^{-1}\left(\frac{y}{\sqrt{1+2y^2}}\right) \end{array}$ Integrating this identity over $[1,\infty)$ , and using the substitution $y = z^{-1}$ , we see that $\begin{array}{rcl} \displaystyle \tfrac1{12}\pi^2 - f(1) \; = \; \int_1^\infty f'(y)\,dy & = & \displaystyle \tfrac{1}{16}\pi^2 - \int_1^\infty \frac{y}{(1+y^2)\sqrt{1+2y^2}} \tan^{-1}\left(\frac{y}{\sqrt{1+2y^2}}\right)\,dy \\ & = & \displaystyle \tfrac{1}{16}\pi^2 - \int_0^1 \frac{1}{(1+z^2)\sqrt{2+z^2}} \tan^{-1}\left(\frac{1}{\sqrt{2+z^2}}\right)\,dz \\ & = & \displaystyle \tfrac{1}{16}\pi^2 - \int_0^1 \frac{1}{(1+z^2)\sqrt{2+z^2}}\left\{\tfrac12\pi - \tan^{-1}\sqrt{2+z^2}\right\}\,dz \\ & = & \displaystyle \tfrac{1}{16}\pi^2 - \tfrac{1}{12}\pi^2 + f(1) \end{array}$ The integral we are interested in is $f(1) \; = \; \tfrac12\left[\tfrac{1}{12}\pi^2 + \tfrac{1}{12}\pi^2 - \tfrac{1}{16}\pi^2\right] \; = \; \tfrac{5}{96}\pi^2$ making the answer $5+2+96 = \boxed{103}$ .

Note: The scribe Ahmes wrote the Rhind Mathematical Papyrus, one of the most important examples of Egyptian mathematics, in about 1650 BC. Changing one letter, the Indian physicist Zafar Ahmed proposed this integral near the start of this century.