If $\mathrm{y}+\frac{\mathrm{d}}{\mathrm{d} x}(x \mathrm{y})=x(\sin x+\log x)$ then

The differential equation $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$ simplifies using the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$.

Substituting gives $y + y + x\frac{dy}{dx} = x(\sin x + \log x)$, which rearranges to $x\frac{dy}{dx} + 2y = x(\sin x + \log x)$.

Dividing by $x$ yields the standard linear form: $\frac{dy}{dx} + \frac{2}{x}y = \sin x + \log x$.

The integrating factor is $e^{\int \frac{2}{x} dx} = x^2$. Multiplying through and integrating gives $y x^2 = \int x^2(\sin x + \log x) dx + C$.

Evaluating the integral by parts: $\int x^2 \sin x dx = -x^2 \cos x + 2x \sin x + 2 \cos x$ and $\int x^2 \log x dx = \frac{x^3}{3} \log x - \frac{x^3}{9}$.

Combining results: $y x^2 = -x^2 \cos x + 2x \sin x + 2 \cos x + \frac{x^3}{3} \log x - \frac{x^3}{9} + C$.

Dividing by $x^2$ produces the solution: $y = -\cos x + \frac{2 \sin x}{x} + \frac{2 \cos x}{x^2} + \frac{x}{3} \log x - \frac{x}{9} + \frac{C}{x^2}$.

This matches option C.