Find the root

The above equation is the same as

$\frac { x+1 }{ \sqrt { { x }^{ 2 }+3 } } +\frac { x+1 }{ \sqrt { 1+3{ x }^{ 2 } } } =2\quad\quad (1)$

Let $a=\frac { x+1 }{ \sqrt { { x }^{ 2 }+3 } } ,\quad b=\frac { x+1 }{ \sqrt { 1+3{ x }^{ 2 } } }$ then $a>0,\quad b>0$ .

We also notice that

${ a }^{ 2 }+{ b }^{ 2 }-2=\frac { { \left( x+1 \right) }^{ 2 } }{ { x }^{ 2 }+3 } +\frac { { \left( x+1 \right) }^{ 2 } }{ 1+3{ x }^{ 2 } } -2=\frac { -2{ \left( x-1 \right) }^{ 4 } }{ \left( { x }^{ 2 }+3 \right) \left( 1+3{ x }^{ 2 } \right) } \le 0\\ \Longrightarrow { a }^{ 2 }+{ b }^{ 2 }\le 2\Longrightarrow { \left( a+b \right) }^{ 2 }\le 2\left( { a }^{ 2 }+{ b }^{ 2 } \right) \le 4\\ \Longrightarrow 0<a+b\le 2.\\ \Longrightarrow LHS\left( 1 \right) \le RHS\left( 1 \right)$

The equality holds when $x=1$ . Hence, $\boxed { 1 }$ is the only root of the provided equation.