How to evaluate Ahmed's integral? How to show that:
$$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$$
I saw this on Wolfram.
 A: A few ways to evaluate it can be found here
Zafar Ahmed, Knut Dale, George Lamb: Definitely an Integral: 10884. The American Mathematical Monthly 109(7): 670-671 (2002)
http://www.jstor.org/stable/pdfplus/3072448.pdf
A: Define $f\left(t\right):=\int_{0}^{1}\frac{\arctan\left(t\sqrt{x^{2}+2}\right)}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}dx$ so$$\begin{align}f\left(\infty\right)&=\frac{\pi}{2}\int_{0}^{1}\frac{dx}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}\\&=\frac{\pi}{2}\left[\arctan\frac{x}{\sqrt{x^{2}+2}}\right]_{0}^{1}\\&=\frac{\pi^{2}}{12},\\f^\prime\left(t\right)&=\frac{1}{1+t^{2}}\int_{0}^{1}\left(\frac{1}{x^{2}+1}-\frac{t^{2}}{1+t^{2}\left(x^{2}+2\right)}\right)dx\\&=\frac{1}{1+t^{2}}\left(\frac{\pi}{4}-\frac{t}{\sqrt{1+2t^{2}}}\arctan\frac{t}{\sqrt{1+2t^{2}}}\right).\end{align}$$Substituting $u=t^{-1}$ and using $\arctan\frac{1}{\theta}=\frac{\pi}{2}-\arctan\theta$, Ahmed's integral $A:=f\left(1\right)$ satisfies$$\frac{\pi^{2}}{12}-A=f\left(\infty\right)-f\left(1\right)=\frac{\pi^{2}}{16}-\left(\frac{\pi^{2}}{12}-A\right)\implies A=\frac{5\pi^{2}}{96}.$$
