∫12x2−1(x4+3x2+1)Tan−1(x2+1x)dx is equal to

Put x+1x=t⇒(1−1x2)dx=dt⇒∫121−1x2(x2+1x2+3).Tan−1(x2+1x)dx={log|Tan−1(x2+1x)|}12=log(Tan−1(52))−log(Tan−12)

∫12x2−1(x4+3x2+1)Tan−1(x2+1x)dx is equal to

Put x+1x=t⇒(1−1x2)dx=dt⇒∫121−1x2(x2+1x2+3).Tan−1(x2+1x)dx={log|Tan−1(x2+1x)|}12=log(Tan−1(52))−log(Tan−12)

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$\int_{1}^{2} \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) T a n^{- 1} (\frac{x^{2} + 1}{x})} d x$ is equal to

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$\log (T a n^{- 1} \frac{5}{2}) - \log (T a n^{- 1} 3)$

$\log (T a n^{- 1} \frac{5}{2}) - \log (T a n^{- 1} 2)$

$\log (T a n^{- 1} \frac{5}{2}) - \log (T a n^{- 1} (\frac{2}{5}))$

$\log (T a n^{- 1} 2) - \log (T a n^{- 1} \frac{2}{5})$

answer is D.

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Detailed Solution

Put $x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t \Rightarrow \int_{1}^{2} \frac{1 - \frac{1}{x^{2}}}{(x^{2} + \frac{1}{x^{2}} + 3) . T a n^{- 1} (\frac{x^{2} + 1}{x})} d x = {\log | T a n^{- 1} (\frac{x^{2} + 1}{x}) |}_{1}^{2}$

$= \log (T a n^{- 1} (\frac{5}{2})) - \log (T a n^{- 1} 2)$

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∫12x2−1(x4+3x2+1)Tan−1(x2+1x)dx is equal to

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$\int_{1}^{2} \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) T a n^{- 1} (\frac{x^{2} + 1}{x})} d x$ is equal to