∫(x2−1)(x4+3x2+1)Tan−1(x2+1x)dx=

∫(x2−1)(x4+3x2+1)Tan−1(x2+1x)dxPut tan−1(x+1x)=t⇒11+x2+1x21-1x2dx⇒x2-1x4+3x2+1dx∫1tdt=logt+c=logtan-1x+1x+c

∫(x2−1)(x4+3x2+1)Tan−1(x2+1x)dx=

∫(x2−1)(x4+3x2+1)Tan−1(x2+1x)dxPut tan−1(x+1x)=t⇒11+x2+1x21-1x2dx⇒x2-1x4+3x2+1dx∫1tdt=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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$l o g s i n^{- 1} (x - \frac{1}{x}) + c$

$l o g \tan^{- 1} (x - \frac{1}{x}) + c$

$l o g \sin^{- 1} (x + \frac{1}{x}) + c$

$l o g \tan^{- 1} (x + \frac{1}{x}) + c$

answer is D.

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Detailed Solution

$\int \frac{(x^{2} - 1)}{(x^{4} + 3 x^{2} + 1) T a n^{- 1} (\frac{x^{2} + 1}{x})} d x$
Put $\tan^{- 1} (x + \frac{1}{x}) = t \Rightarrow \frac{1}{1 + {(\frac{x^{2} + 1}{x})}^{2}} (1 - \frac{1}{x^{2}}) d x \Rightarrow \frac{x^{2} - 1}{x^{4} + 3 x^{2} + 1} d x$
$\int \frac{1}{t} d t = \log t + c = \log \tan^{- 1} (x + \frac{1}{x}) + c$