If ∫x2−1dxx4+3x2+1tan−1x2+1x=logtan−1f(x)+C, then

I=∫x2−1dxx4+3x2+1tan−1x2+1x divide Nr and Dr by x2 ⇒I=∫1−1x2dxx2+3+1x2tan−1x+1x =∫dtt2+1tan−1t · where t=x+1x⇒dt=1−1x2dx =logtan−1t+C =logtan−1x2+1x+C ⇒f(x)=x2+1x

If ∫x2−1dxx4+3x2+1tan−1x2+1x=logtan−1f(x)+C, then

I=∫x2−1dxx4+3x2+1tan−1x2+1x divide Nr and Dr by x2 ⇒I=∫1−1x2dxx2+3+1x2tan−1x+1x =∫dtt2+1tan−1t · where t=x+1x⇒dt=1−1x2dx =logtan−1t+C =logtan−1x2+1x+C ⇒f(x)=x2+1x

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$If \int \frac{(x^{2} - 1) dx}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (\frac{x^{2} + 1}{x})} = \log |\tan^{- 1} f (x)| + C, then$

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$f (x) = x^{2} + 1$

$f (x) = \frac{x^{2} + 1}{2 x}$

$f (x) = \frac{x^{2} + 1}{x}$

$f (x) = \frac{1}{2} (x^{2} + 1)$

answer is C.

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Detailed Solution

$\begin{matrix} I = \int \frac{(x^{2} - 1) d x}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (\frac{x^{2} + 1}{x})} \\ ({divide Nr and Dr by x}^{2}) \\ \Rightarrow I = \int \frac{(1 - \frac{1}{x^{2}}) d x}{(x^{2} + 3 + \frac{1}{x^{2}}) \tan^{- 1} (x + \frac{1}{x})} \\ = \int \frac{d t}{(t^{2} + 1) \tan^{- 1} t} \cdot (where t = x + \frac{1}{x} \Rightarrow d t = (1 - \frac{1}{x^{2}}) d x) \\ = \log |\tan^{- 1} t| + C \\ = \log |\tan^{- 1} (\frac{x^{2} + 1}{x})| + C \\ \Rightarrow f (x) = \frac{x^{2} + 1}{x} \end{matrix}$