Show that $f(x) = 2x + {\cot ^{ - 1}}x + \log \left( {\sqrt {1 + {x^2}} - x} \right)$ is increasing in $R$.

We have, $f(x) = 2x + {\cot ^{ - 1}}x + \log \left( {\sqrt {1 + {x^2}} - x} \right)$

Therefore,${f^\prime }(x) = 2 + \left( {\frac{{ - 1}}{{1 + {x^2}}}} \right) + \frac{1}{{\left( {\sqrt {1 + {x^2}} - x} \right)}}\left( {\frac{1}{{2\sqrt {1 + {x^2}} }} \cdot 2x - 1} \right)$

$= 2 - \frac{1}{{1 + {x^2}}} + \frac{1}{{\left( {\sqrt {1 + {x^2}} - x} \right)}} \cdot \frac{{\left( {x - \sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}$

$= 2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }}$
$= \frac{{2 + 2{x^2} - 1 - \sqrt {1 + {x^2}} }}{{1 + {x^2}}} = \frac{{1 + 2{x^2} - \sqrt {1 + {x^2}} }}{{1 + {x^2}}}$

For the function to be increasing , ${f^\prime }(x) \ge 0$
$\Rightarrow$ $\frac{{1 + 2{x^2} - \sqrt {1 + {x^2}} }}{{1 + {x^2}}} \ge 0$

$\Rightarrow$ $1 + 2{x^2} \ge \sqrt {1 + {x^2}}$

$\Rightarrow$ ${\left( {1 + 2{x^2}} \right)^2} \ge 1 + {x^2}$

$\Rightarrow$ $1 + 4{x^4} + 4{x^2} \ge 1 + {x^2}$

$\Rightarrow$ $4{x^4} + 3{x^2} \ge 0$

$\Rightarrow$ ${x^2}\left( {4{x^2} + 3} \right) \ge 0$

which is true for any real value of $x$.

Hence, $f(x)$ is increasing in $R$.