[-/4 Points] DETAILS Show that $G(x)=x \sin x+\cos x+C$ is the general antiderivative of $g(x)=x \cos x$. \[ \begin{aligned} \frac{d}{d x}(G(x)) & =1 \cdot \square+x \cdot \square-\sin x \\ & =\square+x-\sin x \\ & =x \cos x \end{aligned} \]

Step 1 ：Given the function $G(x)=x \sin x+\cos x+C$, we need to prove that it is the general antiderivative of $g(x)=x \cos x$.

Step 2 ：To do this, we take the derivative of $G(x)$ using the product rule and the chain rule.

Step 3 ：The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Step 4 ：The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 5 ：Applying these rules, we find that the derivative of $G(x)$ is indeed $g(x) = x \cos x$.

Step 6 ：This confirms that $G(x)$ is the antiderivative of $g(x)$.

Step 7 ：Final Answer: Therefore, we have shown that $G(x)=x \sin x+\cos x+C$ is the general antiderivative of $g(x)=x \cos x$. \(\boxed{G(x)=x \sin x+\cos x+C}\)