\(\int[\sin (\log x)+\cos (\log x)] d x=\underline{ }+c\) (a) \(\sin (\log x)+\cos (\log x)\) (b) \(+x \sin (\log x)\) (c) \(-x \cos (\log x)\) (d) \(\sin (\log x)+\cos (\log x)\)

To solve the given integral, we will use integration by parts, where we will consider the following parts: \(u = \sin(\log x) + \cos(\log x)\) and \(dv = dx\) Now, we need to differentiate u and integrate dv: \(du = (\frac{1}{x} \cos(\log x) - \frac{1}{x} \sin(\log x))dx\) \(v = x\)

Using the integration by parts formula, which is: \(\int u dv = uv - \int v du\) We can rewrite the given integral as: \(\int(\sin(\log x) + \cos(\log x))dx = (x[\sin(\log x) + \cos(\log x)]) - \int(x[\frac{1}{x} \cos(\log x) - \frac{1}{x} \sin(\log x)])dx\)

Now, we will simplify the expression before integrating it: \((x[\sin(\log x) + \cos(\log x)]) - \int([x\frac{1}{x} \cos(\log x)] - [x\frac{1}{x} \sin(\log x)])dx\) This simplifies to: \((x[\sin(\log x) + \cos(\log x)]) - \int(\cos(\log x) - \sin(\log x))dx\)

Finally, let's integrate the simplified expression: \((x[\sin(\log x) + \cos(\log x)]) - [\int \cos(\log x)dx - \int \sin(\log x)dx]\) Now we integrate each term separately: \(\int \cos(\log x)dx = x\sin(\log x) + C_1\) \(\int \sin(\log x)dx = -x\cos(\log x) + C_2\) So, the final solution is: \((x[\sin(\log x) + \cos(\log x)]) - [x\sin(\log x) - (-x\cos(\log x))] + C\) \(= x[\sin(\log x) + \cos(\log x) - \sin(\log x) + \cos(\log x)] + C\) This can be further simplified to: \(x[2\cos(\log x)] + C\) Now compare this result with the given options: (a) \(\sin(\log x) + \cos(\log x)\) - Incorrect (b) \(+x \sin(\log x)\) - Incorrect (c) \(-x \cos(\log x)\) - Incorrect (d) \(\sin(\log x) + \cos(\log x)\) - Incorrect No given option matches the obtained result. Therefore, none of the given options (a), (b), (c), and (d) are correct for this integration problem.