∫$$x2$$−$$1x4+3x2+1tan$$−$$1$$⁡$$x+1xdx$$ is equal to

Correct option is 4log⁡tan−1⁡x+1x+C

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$\int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} dx$ is equal to

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tan−1⁡x+1x+C

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detailed solution

Correct option is D

LetI=∫x2−1x4+3x2+1tan−1⁡x+1xdx=∫1−1x2x2+1x2+3tan−1⁡x+1xdxput, x+1x=t⇒1−1x2dx=dtand x+1x2=t2⇒x2+1x2=t2−2∴I=∫1t2+1tan−1⁡tdt=log⁡tan−1⁡t+C =log⁡tan−1⁡x+1x+C

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∫x2−1x4+3x2+1tan−1⁡x+1xdx is equal to

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$\int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} dx$ is equal to