$\int \frac{1}{\sin x-\cos x+\sqrt{2}} d x$ equals

$-\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)$

We have,

$I =\int \frac{1}{\sin x-\cos x+\sqrt{2}} d x$

$\Rightarrow I =\frac{1}{\sqrt{2}} \int \frac{1}{\left(\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x\right)+1} d x$

$\Rightarrow I =\frac{1}{\sqrt{2}} \int \frac{1}{1-\cos (x+\pi / 4)} d x$

$\Rightarrow I =\frac{1}{2 \sqrt{2}} \int ~cosec^2\left(\frac{x}{2}+\frac{\pi}{8}\right) d x$

$\Rightarrow I =\frac{1}{\sqrt{2}} \int ~cosec^2\left(\frac{x}{2}+\frac{\pi}{8}\right) d\left(\frac{x}{2}+\frac{\pi}{8}\right)$

$=-\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)+C$