I=∫xdx2−x2+2−x2; Put t=2−x2 ⇒dtdx=122−x2(−2x)⇒−tdt=xdx∴I=∫(−t)dtt2+t=−∫1t+1dx=−log⁡|t+1|=−log⁡2−x2+1+C

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

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$\log |1 + \sqrt{2 + x^{2}}| + c$

$- \log |1 + \sqrt{2 - x^{2}}| + c$

$- x \log |1 - \sqrt{2 - x^{2}}| + c$

$x \log |1 - \sqrt{2 + x^{2}}| + c$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x d x}{(2 - x^{2}) + \sqrt{2 - x^{2}}}; Put t = \sqrt{2 - x^{2}} \\ \Rightarrow \frac{d t}{d x} = \frac{1}{2 \sqrt{2 - x^{2}}} (- 2 x) \Rightarrow - t d t = x d x \\ ∴ I = \int \frac{(- t) d t}{t^{2} + t} = - \int \frac{1}{t + 1} d x = - \log | t + 1 | \\ = - \log |\sqrt{2 - x^{2}} + 1| + C \end{array}$