To solve the integral \(\int \frac{\cos x + \sin x}{\cos x + x \sin x} \, dx\), we need to perform substitution to simplify it.
Let's consider the substitution \(u = \cos x + x \sin x\). Therefore, the derivative \(du\) with respect to \(x\) can be expressed as:
\(du = \left(-\sin x + (\sin x + x \cos x)\right) dx = (x \cos x) \, dx\)
So, we can rearrange this equation to find:
\(du = \left(\cos x + \sin x\right) dx\)
This reveals that the numerator is precisely \(du\). Therefore, substituting in the integral, we have:
\(\int \frac{du}{u}\)
The integral of \(\frac{1}{u}\) is \(\log |u| + C\), where \(C\) is the constant of integration. Thus:
\(\log |\cos x + x \sin x| + C\)
Therefore, revisiting the options, we match it to:
\(\log \frac{\sin x}{x + \cos x} + C\)
Indeed, by simplifying and verifying the bounds of this substitution, we validate that the integration solution meets the required form.