$If I = \int \frac{\sqrt{1 + x^{2}}}{1 - x^{2}} d x then I is equal to$

Moderate

A

$\frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| + \log |x + \sqrt{1 + x^{2}}| + C$
B

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

$\frac{- 1}{\sqrt{2}} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| - \log |x + \sqrt{1 + x^{2}}| + C$
D

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

Solution

$I = \int \frac{\sqrt{1 + x^{2}}}{1 - x^{2}} \sqrt{\frac{1 + x^{2}}{1 + x^{2}}} d x = \int \frac{(1 + x^{2}) d x}{(1 - x^{2}) \sqrt{1 + x^{2}}}$
$\begin{aligned} I = \int \frac{1 d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} + \int \frac{x^{2} d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \\ I = \int \frac{d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} - \int \frac{- x^{2} d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \end{aligned}$
$\begin{array}{l} I = \int \frac{d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} - \int \frac{[(1 - x^{2}) - 1] d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \\ I = \int \frac{1}{(1 - x^{2}) \sqrt{1 + x^{2}}} d x - \int \frac{(1 - x^{2}) d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} + \int \frac{1 d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \\ I = 2 \int \frac{d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} - \int \frac{d x}{\sqrt{1 + x^{2}}} \\ I = 2 \int \frac{d x}{x^{2} (\frac{1}{x^{2}} - 1) x \sqrt{1 + \frac{1}{x^{2}}}} - \log |x + \sqrt{1 + x^{2}}| \end{array}$
$Put \frac{1}{x^{2}} + 1 = t^{2} in 1^{st} integral$
$\begin{array}{l} - \frac{2}{x^{3}} d x = d t 2 t \\ \frac{d x}{x^{3}} = - t d t \\ = I = 2 \int \frac{- t d t}{(t^{2} - 2) t} - \log |x + \sqrt{x^{2} + 1}| \\ = I = 2 \int \frac{t d t}{(\sqrt{2})^{2} - t^{2}} - \log |x + \sqrt{x^{2} + 1}| + C \\ = I = 2 \cdot \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{2} + t}{\sqrt{2} - t}| - \log |x + \sqrt{1 + x^{2}}| + C \\ = I = \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} + \sqrt{\frac{1}{x^{2}} + 1}}{\sqrt{2} - \sqrt{\frac{1}{x^{2}} + 1}}| - \log |x + \sqrt{1 + x^{2}}| + C \end{array}$
$= I = \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| - \log |x + \sqrt{1 + x^{2}}| + C$

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