∫x2−1x4+3x2+1Tan−1⁡x2+1xdx=

∫x2−1x4+3x2+1tan−1⁡x2+1xdx=∫x2−1x2x2+3+1x2tan−1⁡x+1x =∫1−1/x2x2+1x2+3tan−1⁡x+1x put x+1x=t⇒1-1x2dx=dt=∫dtt2+1tan−1⁡(t) =∫dtt2+1tan−1⁡tdt=log⁡tan−1⁡t+c=log⁡tan−1⁡x+1x+c

∫x2−1x4+3x2+1Tan−1⁡x2+1xdx=

∫x2−1x4+3x2+1tan−1⁡x2+1xdx=∫x2−1x2x2+3+1x2tan−1⁡x+1x =∫1−1/x2x2+1x2+3tan−1⁡x+1x put x+1x=t⇒1-1x2dx=dt=∫dtt2+1tan−1⁡(t) =∫dtt2+1tan−1⁡tdt=log⁡tan−1⁡t+c=log⁡tan−1⁡x+1x+c

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

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$\log \sin^{- 1} (x + \frac{1}{x}) + c$

$\log \sin^{- 1} (x - \frac{1}{x}) + c$

$\log {Tan}^{- 1} (x + \frac{1}{x}) + c$

$\log {Tan}^{- 1} (x - \frac{1}{x}) + c$

answer is D.

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Detailed Solution

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

$\begin{array}{l} = \int \frac{x^{2} - 1}{x^{2} (x^{2} + 3 + \frac{1}{x^{2}}) \tan^{- 1} (x + \frac{1}{x})} \\ = \int \frac{1 - 1 / x^{2}}{(x^{2} + \frac{1}{x^{2}} + 3) \tan^{- 1} (x + \frac{1}{x})} p u t x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t \\ = \int \frac{dt}{(t^{2} + 1) \tan^{- 1} (t)} \end{array}$

$\begin{array}{l} = \int \frac{\frac{dt}{t^{2} + 1}}{\tan^{- 1} t} dt \\ = \log (\tan^{- 1} t) + c \\ = \log (\tan^{- 1} (x + \frac{1}{x})) + c \end{array}$