1. We need to evaluate

$$\sqrt{2}\int \frac{\sin x}{\sin\left(x-\frac{\pi}{4}\right)}\,dx.$$

2. Use the identity

$$\sin\left(x-\frac{\pi}{4}\right)=\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4} =\frac{1}{\sqrt{2}}(\sin x-\cos x).$$

Hence,

$$\sqrt{2}\cdot \sin\left(x-\frac{\pi}{4}\right)=\sin x-\cos x.$$

So the integrand becomes

$$\sqrt{2}\frac{\sin x}{\sin\left(x-\frac{\pi}{4}\right)} =\frac{\sqrt{2}\sin x}{\sin\left(x-\frac{\pi}{4}\right)} =\frac{2\sin x}{\sin x-\cos x}.$$

Thus,

$$\sqrt{2}\int \frac{\sin x}{\sin\left(x-\frac{\pi}{4}\right)}dx =\int \frac{2\sin x}{\sin x-\cos x}\,dx.$$

3. Rewrite the numerator:

$$2\sin x=(\sin x-\cos x)+(\sin x+\cos x).$$

Therefore,

$$\frac{2\sin x}{\sin x-\cos x} =1+\frac{\sin x+\cos x}{\sin x-\cos x}.$$

So the integral becomes

$$\int \frac{2\sin x}{\sin x-\cos x}\,dx =\int 1\,dx+\int \frac{\sin x+\cos x}{\sin x-\cos x}\,dx.$$

4. For the second integral, let

$$u=\sin x-\cos x.$$

Then

$$\frac{d\nu}{dx}=\cos x+\sin x,$$

so

$$d\nu=(\sin x+\cos x)dx.$$

Hence,

$$\int \frac{\sin x+\cos x}{\sin x-\cos x}\,dx =\int \frac{d\nu}{\nu} =\log|\nu|+C =\log|\sin x-\cos x|+C.$$

5. Therefore,

$$\sqrt{2}\int \frac{\sin x}{\sin\left(x-\frac{\pi}{4}\right)}dx =x+\log|\sin x-\cos x|+C.$$

Now use

$$\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right).$$

Thus,

$$\log|\sin x-\cos x|= \log\left|\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\right| =\log\sqrt{2}+\log\left|\sin\left(x-\frac{\pi}{4}\right)\right|.$$

Since $\log\sqrt{2}$ is a constant, it gets absorbed into $C$.

So the final result is

$$\boxed{x+\log\left|\sin\left(x-\frac{\pi}{4}\right)\right|+C}.$$

6. Checking options:

• A: involves $\log|\cos(x-\pi/4)|$ — incorrect
• B: wrong sign before log — incorrect
• C: $x+\log\left|\sin\left(x-\frac{\pi}{4}\right)\right|+c$ — correct
• D: wrong function and wrong sign — incorrect