∫(x2−1)x4+3x2+1Tan−1x2+1xdx=

∫(x2−1)x4+3x2+1Tan−1x2+1xdx=∫x2−1x2x2+3+1x2tan−1x2+1xdx =∫1-1x2dxx+1x2+1tan-1x+1x Put tan-1x+1x=t 11+x+1x21-1x2dx=dt =∫dtt=logt+C =logtan-1x+1x+C

∫(x2−1)x4+3x2+1Tan−1x2+1xdx=

∫(x2−1)x4+3x2+1Tan−1x2+1xdx=∫x2−1x2x2+3+1x2tan−1x2+1xdx =∫1-1x2dxx+1x2+1tan-1x+1x Put tan-1x+1x=t 11+x+1x21-1x2dx=dt =∫dtt=logt+C =logtan-1x+1x+C

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

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

$\log T a n^{- 1} (x + \frac{1}{x}) + c$

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

$\log T a n^{- 1} (x - \frac{1}{x}) + c$

answer is D.

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Detailed Solution

$\int \frac{(x^{2} - 1)}{(x^{4} + 3 x^{2} + 1) T a n^{- 1} (\frac{x^{2} + 1}{x})} d x$

$= \int \frac{x^{2} - 1}{x^{2} (x^{2} + 3 + \frac{1}{x^{2}}) t a n^{- 1} (\frac{x^{2} + 1}{x})} d x = \int \frac{(1 - \frac{1}{x^{2}}) d x}{[{(x + \frac{1}{x})}^{2} + 1] \tan^{- 1} (x + \frac{1}{x})} P u t \tan^{- 1} (x + \frac{1}{x}) = t \frac{1}{1 + {(x + \frac{1}{x})}^{2}} (1 - \frac{1}{x^{2}}) d x = d t = \int \frac{d t}{t} = \log |t| + C = \log |\tan^{- 1} (x + \frac{1}{x})| + C$