$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx =$

(Given that $c$ is an arbitrary constant)

Using the double angle formula $\cos 2\theta = 2\cos^2\theta - 1$:

$\cos 2x = 2\cos^2 x - 1$

$\cos 2\alpha = 2\cos^2 \alpha - 1$

The numerator becomes:

$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1)$

$= 2\cos^2 x - 2\cos^2 \alpha$

$= 2(\cos^2 x - \cos^2 \alpha)$

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Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$:

$\cos^2 x - \cos^2 \alpha = (\cos x - \cos \alpha)(\cos x + \cos \alpha)$

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The fraction becomes:

$\frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha}$

Canceling the common term $(\cos x - \cos \alpha)$:

$= 2(\cos x + \cos \alpha)$

$= 2\cos x + 2\cos \alpha$

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The integral becomes:

$\int (2\cos x + 2\cos \alpha) \, dx$

Since $\alpha$ is a constant:

$\int 2\cos x \, dx = 2\sin x$

$\int 2\cos\alpha \, dx = 2x\cos\alpha$

Therefore:

$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx = 2\sin x + 2x\cos\alpha + c$