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$\int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} d x =$

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tan−1⁡x+1x+C

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ln⁡tan−1⁡x+1x+C

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detailed solution

Correct option is C

I=∫x2−1x4+3x2+1tan−1⁡x+1xdx =∫1−1x2x2+1x2+3tan−1⁡x+1xdx Put x+1x=t⇒1−1x2dx=dt and x2+1x2+2=t2 I=∫dtt2+1tan−1⁡t=ln⁡tan−1⁡t+c =ln⁡tan−1⁡x+1x+c

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∫x2−1x4+3x2+1tan−1⁡x+1xdx=

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$\int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} d x =$