f(x)=2x−Tan−1⁡x−log⁡x+1+x2(x>0) is increasing in

f1(x)>0⇒2x2+1−1+x2>0⇒x24x2+3>0⇒x>0

f1(x)>0⇒2x2+1−1+x2>0⇒x24x2+3>0⇒x>0

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$f (x) = 2 x - {Tan}^{- 1} x - \log (x + \sqrt{1 + x^{2}}) (x > 0)$ is increasing in

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$(0, 1) \cup (2, \infty)$

$(1, 2)$

$(0, \infty)$

$(- \infty, \infty)$

answer is C.

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$f^{1} (x) > 0 \Rightarrow (2 x^{2} + 1) - \sqrt{1 + x^{2}} > 0$

$\Rightarrow x^{2} (4 x^{2} + 3) > 0 \Rightarrow x > 0$

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