# log x -tan^-1 x increases in the interval

Solution:   We have ,     $f(x)=\log_{e}x -\tan^{-1} x$

$\Rightarrow$       $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{x}-\frac{1}{1+x^{2}}=\frac{1+x^{2}-x}{x(1+x^{2})}$

$\Rightarrow$                         $=\frac{(x-\frac{1}{2})^{2}+\frac{3}{4}}{x(1+x^{2})}> 0$    $\forall$bbb$x> 0$

Hence ,   $f(x)$   is  an increasing function in the interval   $(0,\infty )$

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