∫x2−1dxx4+3x2+1tan−1⁡x2+1x=ln⁡(f(x))+c then f(x)=

Let I=∫x2−1dxx4+3x2+1tan−1⁡x+1x =∫1−1x2dxx2+3+1x2tan−1⁡x+1x=∫1−1x2dxx+1x2+1tan−1⁡x+1x Put x+1x=t⇒1−1x2dx=dt I=∫dtt2+1tan−1⁡t Now put tan−1⁡t=u ⇒dt1+t2=du I=∫duuI=log⁡u+C=log⁡tan−1⁡t+C⇒I=log⁡tan−1⁡x+1x+C

∫x2−1dxx4+3x2+1tan−1⁡x2+1x=ln⁡(f(x))+c then f(x)=

Let I=∫x2−1dxx4+3x2+1tan−1⁡x+1x =∫1−1x2dxx2+3+1x2tan−1⁡x+1x=∫1−1x2dxx+1x2+1tan−1⁡x+1x Put x+1x=t⇒1−1x2dx=dt I=∫dtt2+1tan−1⁡t Now put tan−1⁡t=u ⇒dt1+t2=du I=∫duuI=log⁡u+C=log⁡tan−1⁡t+C⇒I=log⁡tan−1⁡x+1x+C

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$\int \frac{(x^{2} - 1) dx}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (\frac{x^{2} + 1}{x})} = \ln (f (x)) + c then f (x) =$

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

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

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

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

answer is B.

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Detailed Solution

$Let I = \int \frac{(x^{2} - 1) d x}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} \begin{aligned} = \int \frac{(1 - \frac{1}{x^{2}}) d x}{(x^{2} + 3 + \frac{1}{x^{2}}) \tan^{- 1} (x + \frac{1}{x})} \\ = \int \frac{(1 - \frac{1}{x^{2}}) d x}{({(x + \frac{1}{x})}^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} \end{aligned} Put x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t I = \int \frac{d t}{(t^{2} + 1) \tan^{- 1} t} Now put \tan^{- 1} t = u \Rightarrow \frac{dt}{1 + t^{2}} = du \begin{array}{l} I = \int \frac{du}{u} \\ I = \log u + C \\ = \log \tan^{- 1} t + C \\ \Rightarrow I = \log \tan^{- 1} (x + \frac{1}{x}) + C \end{array}$