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?
 A: We have $$I=\int_{0}^{\infty}\frac{\sqrt{x^{2}+1}+x^{2}\sqrt{x^{2}+2}}{\sqrt{\left(x^{2}+1\right)\left(x^{2}+2\right)}}\frac{1}{\left(x^{2}+1\right)^{2}}dx=\int_{0}^{\infty}\frac{x^{2}}{\left(x^{2}+1\right)^{5/2}}dx+\int_{0}^{\infty}\frac{1}{\left(x^{2}+1\right)^{2}\sqrt{x^{2}+2}}dx=I_{1}+I_{2}.
 $$ Now recalling that the Beta function has the representation $$\int_{0}^{\infty}\frac{u^{m}}{\left(u+1\right)^{m+n+2}}du=B\left(m+1,n+1\right)
  $$ we get $$I_{1}=\frac{1}{2}\int_{0}^{\infty}\frac{u^{1/2}}{\left(u+1\right)^{5/2}}du=\frac{B\left(\frac{3}{2},1\right)}{2}=\frac{1}{3}.
 $$ If you prefer, you can also compute $I_{1}
 $ using the substitution $x=\tan\left(v\right)
 $. For $I_{2}
 $ we get $$I_{2}\stackrel{x=\sqrt{2}\tan\left(v\right)}{=}\int_{0}^{\pi/2}\frac{\left(1-\sin^{2}\left(v\right)\right)\cos\left(v\right)}{\left(\sin^{2}\left(v\right)+1\right)^{2}}dv
 $$ $$\stackrel{z=\sin\left(v\right)}{=}\int_{0}^{1}\frac{1-z^{2}}{\left(z^{2}+1\right)^{2}}dz=\int_{0}^{1}\frac{2}{\left(z^{2}+1\right)^{2}}dz-\int_{0}^{1}\frac{1}{z^{2}+1}dz
 $$ and note that $$\int_{0}^{1}\frac{2}{\left(z^{2}+1\right)^{2}}dz\stackrel{z=\tan\left(w\right)}{=}2\int_{0}^{\pi/4}\cos^{2}\left(w\right)dw=\frac{1}{2}+\frac{\pi}{4}
 $$ hence $$I=\frac{1}{3}+\frac{\pi}{4}+\frac{1}{2}-\frac{\pi}{4}=\color{red}{\frac{5}{6}}.$$
A: Let $A=\sqrt{x^2+1}$ and $B=\sqrt{x^2+2}$ so that have $A^2-1=x^2$. With these you have 
\begin{align}
{\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}&=\frac{A+(A^2-1)B}{AB}\frac{1}{A^4}\\
&=\frac{1}{BA^4}-\frac{1}{A^3}-\frac{1}{A^5}\\
\end{align}
Note that the integral of all of these terms is very simple to evaluate, so have fun :-)
A: on primitive function should be $${\frac {\sqrt {{x}^{2}+1}\sqrt {{x}^{2}+2}}{\sqrt { \left( {x}^{2}+1
 \right)  \left( {x}^{2}+2 \right) }}\arctan \left( {\frac {x}{\sqrt {
{x}^{2}+2}}} \right) }+{\frac {\sqrt {{x}^{2}+2} \left( {\rm arcsinh} 
\left(x\right)\sqrt {{x}^{2}+1}-x \right) }{\sqrt { \left( {x}^{2}+1
 \right)  \left( {x}^{2}+2 \right) }}}
$$
A: $u=(x^2+1)^{\frac{1}{2}}$, thus $du = \frac{1}{2}(x^2+1)^{-\frac{1}{2}}2xdx
=\frac{x}{(x^2+1)^{\frac{1}{2}}}dx$.
$x \in (0, +\infty), u \in (1, +\infty), x =(u^2-1)^{\frac{1}{2}}$
$dx = \frac{(x^2+1)^{1/2}}{x}du=\frac{u}{(u^2-1)^{1/2}}du$
$\int_0^\infty\frac{u+(u^2-1)(u^2+1)^{1/2}}{u^2(u^2+1)^{1/2}} \cdot \frac{1}{u^4} \cdot \frac{u}{(u^2-1)^{1/2}}du$
I find some mistakes, but I can not solve it. Sorry for that.
