Help to prove that $\int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$ I am trying to prove that
$$\displaystyle \int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$$
$u=(x^2+1)^{1/2}$ then $du=(x^2+1)^{-1/2}dx$
$$\int_{1}^{\infty}{u^2+(u^2-1)\sqrt{u^2+1}\over u^2\sqrt{u^2+1}}du$$
$v=(u^2+1)^{1/2}$ then $dv=(u^2+1)^{-1/2}du$
$$\int_{1}^{\infty}{v^3+v^2-2v-1\over v^2-1}dv$$
$\int_{1}^{\infty}{v^2\over v^2-1}-{1\over v^2-1}dv$ -$\ln{(v^2-1)}|_{1}^{\infty}$
I am sure I when wrong somewhere, but I can figured it out.
Any help?

on primitive function should be $${\frac {\sqrt {{x}^{2}+1}\sqrt {{x}^{2}+2}}{\sqrt { \left( {x}^{2}+1
 \right)  \left( {x}^{2}+2 \right) }}\arctan \left( {\frac {x}{\sqrt {
{x}^{2}+2}}} \right) }+{\frac {\sqrt {{x}^{2}+2} \left( {\rm arcsinh} 
\left(x\right)\sqrt {{x}^{2}+1}-x \right) }{\sqrt { \left( {x}^{2}+1
 \right)  \left( {x}^{2}+2 \right) }}}
$$
