∫x3+x2+x−1x2−12−x2dx is equal to

∫x3+x2+x−1(x2−1)2−x2dx=∫x2−1+x3+x(x2−1)2−x2dx=∫12−x2dx+∫xx2+1x2−12−x2 Let 2−x2=t22xdx=−2tdt=sin−1⁡x2−∫t3−t21−t2tdt=sin−1⁡x2−∫1−t2+21−t2=sin−1⁡x2−∫1+21−t2dt=sin−1⁡x2−t+212(1)log⁡1+t1−t+C=sin−1⁡x2−2−x2+log⁡1+2−x21−2−x2+C=sin−1⁡x2−2−x2−log⁡1+2−x21−2−x2+C

∫x3+x2+x−1x2−12−x2dx is equal to

∫x3+x2+x−1(x2−1)2−x2dx=∫x2−1+x3+x(x2−1)2−x2dx=∫12−x2dx+∫xx2+1x2−12−x2 Let 2−x2=t22xdx=−2tdt=sin−1⁡x2−∫t3−t21−t2tdt=sin−1⁡x2−∫1−t2+21−t2=sin−1⁡x2−∫1+21−t2dt=sin−1⁡x2−t+212(1)log⁡1+t1−t+C=sin−1⁡x2−2−x2+log⁡1+2−x21−2−x2+C=sin−1⁡x2−2−x2−log⁡1+2−x21−2−x2+C

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$\int \frac{x^{3} + x^{2} + x - 1}{(x^{2} - 1) \sqrt{2 - x^{2}}} d x is equal to$

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$\sin^{- 1} (\frac{x}{\sqrt{2}}) + \sqrt{2 - x^{2}} + \log |\frac{1 + \sqrt{2 - x^{2}}}{1 - \sqrt{2 - x^{2}}}| + C$

$\sin^{- 1} (\frac{x}{\sqrt{2}}) - \sqrt{2 - x^{2}} + \log |\frac{1 + \sqrt{2 - x^{2}}}{1 - \sqrt{2 - x^{2}}}| + C$

$\sin^{- 1} (\frac{x}{\sqrt{2}}) + \sqrt{2 - x^{2}} - \log |\frac{1 + \sqrt{2 - x^{2}}}{1 - \sqrt{2 - x^{2}}}| + C$

$\sin^{- 1} (\frac{x}{\sqrt{2}}) - \sqrt{2 - x^{2}} - \log |\frac{1 + \sqrt{2 - x^{2}}}{1 - \sqrt{2 - x^{2}}}| + C$

answer is D.

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Detailed Solution

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

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

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

$Let 2 - x^{2} = t^{2}$

$2 x d x = - 2 t d t$

$\begin{array}{l} = \sin^{- 1} (\frac{x}{\sqrt{2}}) - \int \frac{t [3 - t^{2}]}{(1 - t^{2}) t} d t \\ = \sin^{- 1} (\frac{x}{\sqrt{2}}) - \int \frac{(1 - t^{2}) + 2}{(1 - t^{2})} \\ = \sin^{- 1} (\frac{x}{\sqrt{2}}) - \int 1 + \frac{2}{1 - t^{2}} d t \\ = \sin^{- 1} (\frac{x}{\sqrt{2}}) - [t + 2 \frac{1}{2 (1)} \log |\frac{1 + t}{1 - t}|] + C \\ = \sin^{- 1} (\frac{x}{\sqrt{2}}) - [\sqrt{2 - x^{2}} + \log |\frac{1 + \sqrt{2 - x^{2}}}{1 - \sqrt{2 - x^{2}}}|] + C \\ = \sin^{- 1} (\frac{x}{\sqrt{2}}) - \sqrt{2 - x^{2}} - \log |\frac{1 + \sqrt{2 - x^{2}}}{1 - \sqrt{2 - x^{2}}}| + C \end{array}$