Different approach to tackle $\int_{0}^{\infty}{2\sqrt{x^2+1}+\gamma(x^2-x)\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{\mathrm dx\over (x^2+1)^2}=1$ $$\int_{0}^{\infty}{2\sqrt{x^2+1}+\gamma(x^2-x)\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{\mathrm dx\over (x^2+1)^2}=1\tag1$$
Where $\gamma=0.577...$ it is Euler's Constant
$$\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}+\gamma\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx\tag2$$
$$\gamma\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx=\gamma\int_{0}^{\infty}{x^2\over (x^2+1)^{5/2}}\mathrm dx-\gamma\int_{0}^{\infty}{x\over (x^2+1)^{5/2}}\mathrm dx\tag3$$
Enforcing $u=x^2$ then $du=2xdx$
Recalling from the beta function
$$\int_{0}^{\infty}{u^m\over (u+1)^{m+n+2}}du=B(m+1,n+1)\tag4$$
$${\gamma\over 2}\int_{0}^{\infty}{u\over (u+1)^{5/2}}\mathrm du-{\gamma\over 2}\int_{0}^{\infty}{1\over (u+1)^{5/2}}\mathrm du={B(3/2,1)\over2}-{B(1,3/2)\over2}=0\tag5$$
So this part of integration must be 1! It is ready done my @Marco on my previous post.
$$\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}\tag6$$
Can anyone prove this integral $(1)$ via another method? Thank you! Sorry for not addressing the question properly.

In fact, letting $x=\tan t$ and $u=\sin t$, one has
\begin{eqnarray}
&&\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}\\
&=&2\int_0^{\pi/2}\frac{1}{\sqrt{1+\sec^2t}}\frac{1}{\sec^4t}\sec^2tdt\\
&=&2\int_0^{\pi/2}\frac{\cos^3 t}{\sqrt{1+\sec^2t}}dt\\
&=&2\int_0^{\pi/2}\frac{\cos^3 t}{\sqrt{2-\sin^2t}}dt\\
&=&2\int_0^{1}\frac{1-u^2}{\sqrt{2-u^2}}du\\
&=&u\sqrt{2-u^2}|_0^1\\
&=&1.\\
&&\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\tan^2t-\tan t\over (\sec^2t)^{5/2}}\sec^2t\mathrm dx\\
&=&\int_{0}^{\pi/2}(\tan^2t-\tan t)\cos t\mathrm dx\\
&=&\int_{0}^{\pi/2}\cos t\sin t(\sin t-\cos t)\mathrm dx\\
&=&\int_{0}^{\pi/2}\cos t\sin^2 t\mathrm dx-\int_{0}^{\pi/2}\cos^2 t\sin t\mathrm dx\\
&=&0.
\end{eqnarray}
