∫dxx2−1x2+2=∫131x2−1−1x2+2dx⇒13∫1x2−12dx−13∫1x2+(2)2dx=13⋅12log⁡x−1x+1−13⋅12Tan−1⁡x2+c⇒16log⁡x−1x+1−132Tan−1⁡x2+c=1312log⁡x−1x+1−12Tan′⁡x2+c

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$\int \frac{1}{(x^{2} - 1) (x^{2} + 2)} dx =$

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$\frac{1}{3} [\frac{1}{2} \log |\frac{x + 1}{x - 1}| - \frac{1}{\sqrt{2}} {Tan}^{- 1} \frac{x}{\sqrt{2}}] + c$

$\frac{1}{3} [\frac{1}{2} \log |\frac{x - 1}{x + 1}| + \frac{1}{\sqrt{2}} {Tan}^{- 1} \frac{x}{\sqrt{2}}] + c$

$\frac{1}{3} [\frac{1}{2} \log |\frac{x - 1}{x + 1}| - \frac{1}{\sqrt{2}} {Tan}^{- 1} \frac{x}{\sqrt{2}}] + c$

$\frac{1}{3} [\frac{1}{3} \log |\frac{x - 1}{x + 1}| - \frac{1}{2 \sqrt{2}} {Tan}^{- 1} \frac{x}{\sqrt{2}}] + c$

answer is C.

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Detailed Solution

$\begin{matrix} \int \frac{d x}{(x^{2} - 1) (x^{2} + 2)} = \int \frac{1}{3} [\frac{1}{x^{2} - 1} - \frac{1}{x^{2} + 2}] d x \Rightarrow \frac{1}{3} \int \frac{1}{x^{2} - 1^{2}} d x - \frac{1}{3} \int \frac{1}{x^{2} + (\sqrt{2})^{2}} d x \\ = \frac{1}{3} \cdot \frac{1}{2} \log |\frac{x - 1}{x + 1}| - \frac{1}{3} \cdot \frac{1}{\sqrt{2}} {Tan}^{- 1} (\frac{x}{\sqrt{2}}) + c \Rightarrow \frac{1}{6} \log |\frac{x - 1}{x + 1}| - \frac{1}{3 \sqrt{2}} {Tan}^{- 1} (\frac{x}{\sqrt{2}}) + c \\ = \frac{1}{3} [\frac{1}{2} \log |\frac{x - 1}{x + 1}| - \frac{1}{\sqrt{2}} {Tan}^{'} (\frac{x}{\sqrt{2}})] + c \end{matrix}$