Prove
$$I=\int\_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx} = \frac{{\ln 3}}{2} - \frac{{\ln 2}}{3}.$$

First note that
$$4{x^4} + 8{x^3} + 12{x^2} + 8x + 1 = 4{\left( {{x^2} + x + 1} \right)^2} - 3,$$
we let
$${x^2} + x + 1 = \frac{{\sqrt 3 }}{{2\cos \theta }} \Rightarrow x = \sqrt { - \frac{3}{4} + \frac{{\sqrt 3 }}{{2\cos \theta }}} - \frac{1}{2},$$
then
$$I=\frac{1}{2}\int\_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)\sec \theta }}{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} + 1} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}d\theta } .$$
we have
\begin{align\*}
&\frac{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)\sec \theta }}{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} + 1} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} = \frac{{{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)}^2}\sec \theta }}{{\left( {2\sqrt 3 \sec \theta - 4} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}\\
 =& \frac{{\left( {2\sqrt 3 \sec \theta - 2 - 2\sqrt {2\sqrt 3 \sec \theta - 3} } \right)\sec \theta }}{{\left( {2\sqrt 3 \sec \theta - 4} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} = \frac{{\left( {\sqrt 3 \sec \theta - 1 - \sqrt {2\sqrt 3 \sec \theta - 3} } \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}\\
 = &\frac{{\left( {\sqrt 3 \sec \theta - 1} \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} - \frac{{\sec \theta }}{{\sqrt 3 \sec \theta - 2}}.
\end{align\*}
and
$$\int {\frac{{\sec \theta }}{{\sqrt 3 \sec \theta - 2}}d\theta } = \ln \frac{{\left( {2 + \sqrt 3 } \right)\tan \frac{\theta }{2} - 1}}{{\left( {2 + \sqrt 3 } \right)\tan \frac{\theta }{2} + 1}}+ C.$$
while
\begin{align\*}&\int {\frac{{\left( {\sqrt 3 \sec \theta - 1} \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}d\theta } = \int {\frac{{\sqrt 3 - \cos \theta }}{{\left( {\sqrt 3 - 2\cos \theta } \right)\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } \\
 = &\frac{1}{2}\int {\frac{1}{{\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } + \frac{{\sqrt 3 }}{2}\int {\frac{1}{{\left( {\sqrt 3 - 2\cos \theta } \right)\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } .
\end{align\*}
But how can we continue? It is related to elliptic integral.
['https://math.stackexchange.com/questions/2127460', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/165013/']
This is a pseudo-elliptic integral, it has an elementary anti-derivative:

$$\int \frac{x}{(x+1)\sqrt{4x^4+8x^3+12x^2+8x+1}} dx = \frac{\ln\left[P(x)+Q(x)\sqrt{4x^4+8x^3+12x^2+8x+1}\right]}{6} - \ln(x+1) + C$$

where $$P(x) = 112x^6+360x^5+624x^4+772x^3+612x^2+258x+43$$
and
$$Q(x) = 52x^4+92x^3+30x^2-22x-11$$

To obtain this answer, just follow the systematic method of symbolic integration over simple algebraic extension. Alternatively, you can throw it to a CAS with Risch algorithm implemented (**not Mathematica**), a convenient software is the online **Axiom sandbox**.
**HINT 1**

$$x=2x+1-(x+1),$$
so 
$$I=\frac12\int\_0^\infty\frac{(2x+1)dx}{(x+1)\sqrt{(x^2+x+1)^2-\frac34}} - \frac12\int\_0^\infty\frac{(2x+1)dx}{(2x+1)\sqrt{(x^2+x+1)^2-\frac34}}.$$

**HINT 2**

$$(2x+1)dx = d(x^2+x+1) = \frac{\sqrt3}2d\sec\theta$$