Put x=1t⇒dx=−1t2dt=∫−tt2+1t2−1Put t2−1=yt2=1+y2tddt=ydy=−∫1y2+2dy=−12tan−1⁡y2+c=−12tan−1⁡t2−12+c=−12tan−1⁡1−x2r2+c

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

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$\frac{- 1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{\sqrt{2} x}) + c$

$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{2} x}{\sqrt{1 + x^{2}}}) + c$

$\frac{1}{2} \tan^{- 1} (\frac{\sqrt{2} x}{\sqrt{1 - x^{2}}}) + c$

$\frac{1}{2} \tan^{- 1} (\frac{\sqrt{2} x}{\sqrt{1 + x^{2}}}) + c$

answer is A.

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Detailed Solution

Put $x = \frac{1}{t} \Rightarrow dx = \frac{- 1}{t^{2}} dt$
$= \int \frac{- t}{(t^{2} + 1) \sqrt{t^{2}} - 1}$
Put $\sqrt{t^{2} - 1} = y$
$\begin{array}{l} t^{2} = 1 + y^{2} \\ tddt = ydy \\ = - \int \frac{1}{y^{2} + 2} dy = \frac{- 1}{\sqrt{2}} \tan^{- 1} (\frac{y}{\sqrt{2}}) + c \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{t^{2} - 1}{2}}) + c \\ = \frac{- 1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{r \sqrt{2}}) + c \end{array}$