∫(x2−1)dx(x4+3x2+1)tan−1(x2+1x)=ln|f(x)|+C then f(x) is:

∫1−1x2(x2+1x2+3)tan−1(x+1x)dx; ∫(1−1x2)dx((x+1x)2+1)tan−1(x+1x)dx

∫(x2−1)dx(x4+3x2+1)tan−1(x2+1x)=ln|f(x)|+C then f(x) is:

∫1−1x2(x2+1x2+3)tan−1(x+1x)dx; ∫(1−1x2)dx((x+1x)2+1)tan−1(x+1x)dx

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$\int \frac{(x^{2} - 1) d x}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (\frac{x^{2} + 1}{x})} = \ln | f (x) | + C$ then $f (x)$ is:

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$\ln (x + \frac{1}{x})$

$\tan^{- 1} (x + \frac{1}{x})$

$\ln [\tan^{- 1} (x + \frac{1}{x})]$

$\cot^{- 1} (x + \frac{1}{x})$

answer is B.

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Detailed Solution

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

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