$\int \frac{dx}{\sin x - \cos x + \sqrt{2}}$ is equal to

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$- \frac{1}{\sqrt{2}} \tan (\frac{x}{2} + \frac{π}{8}) + C$

$\frac{1}{\sqrt{2}} \tan (\frac{x}{2} + \frac{π}{8}) + C$

$\frac{1}{\sqrt{2}} \cot (\frac{x}{2} + \frac{π}{8}) + C$

$- \frac{1}{\sqrt{2}} \cot (\frac{x}{2} + \frac{π}{8}) + C$

answer is C.

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Detailed Solution

$\begin{array}{l} \int \frac{dx}{\sin x - \cos x + \sqrt{2}} \\ = \int \frac{dx}{\sin x \frac{\sqrt{2}}{\sqrt{2}} - \cos x \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2}} \\ = \int \frac{dx}{\sqrt{2} (\sin xsin \frac{π}{4} - \cos xcos \frac{π}{4} + 1)} \\ = \frac{1}{\sqrt{2}} \int \frac{dx}{1 - \cos (x + \frac{π}{4})} \\ = \frac{1}{\sqrt{2}} \int \frac{dx}{1 - \cos 2 (\frac{x}{2} + \frac{π}{8})} = \frac{1}{\sqrt{2}} \int \frac{dx}{2 \sin^{2} (\frac{x}{2} + \frac{π}{8})} \\ = \frac{1}{2 \sqrt{2}} \int {cosec}^{2} (\frac{x}{2} + \frac{π}{8}) dx \\ = - \frac{1}{2 \sqrt{2}} \cdot \frac{- \cot (\frac{x}{2} + \frac{π}{8})}{\frac{1}{2}} + C = \frac{1}{\sqrt{2}} \cot (\frac{x}{2} + \frac{π}{8}) + C \end{array}$