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If $$f(x) = 2\tan^{-1} x + \sin^{-1}\left(\frac{2x}{1+x^2}\right)$$, $$x > 1$$, then $$f(5)$$ is equal to
$$f(x) = 2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1+x^2} \right)$$
If $$x > 1$$: $$\sin^{-1} \left( \frac{2x}{1+x^2} \right) = \pi - 2 \tan^{-1} x$$
$$\sin^{-1} \left( \frac{2x}{1+x^2} \right) = \pi - 2 \tan^{-1} x$$
$$f(x) = 2 \tan^{-1} x + \left( \pi - 2 \tan^{-1} x \right)$$
$$f(x) = \pi$$
$$f(5) = \pi$$
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