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(C) Given that the antiderivative of $f(x)$ is $e^x$, we have $\int f(x) \, dx = e^x$, which implies $f(x) = \frac{d}{dx}(e^x) = e^x$.Given that the a

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$e^x \cos x + c$

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(C) Given that the antiderivative of $f(x)$ is $e^x$, we have $\int f(x) \, dx = e^x$, which implies $f(x) = \frac{d}{dx}(e^x) = e^x$.Given that the antiderivative of $g(x)$ is $\cos x$, we have $\int g(x) \, dx = \cos x$, which implies $g(x) = \frac{d}{dx}(\cos x) = -\sin x$.We need to evaluate $I = \int f(x) \cos x \, dx + \int g(x) e^x \, dx$.Substituting the expressions for $f(x)$ and $g(x)$:$I = \int e^x \cos x \, dx + \int (-\sin x) e^x \, dx$.$I = \int e^x \cos x \, dx - \int e^x \sin x \, dx$.Using the product rule for differentiation: $\frac{d}{dx}(e^x \cos x) = e^x \cos x + e^x(-\sin x) = e^x \cos x - e^x \sin x$.Integrating both sides with respect to $x$:$e^x \cos x = \int (e^x \cos x - e^x \sin x) \, dx + c$.$e^x \cos x = \int e^x \cos x \, dx - \int e^x \sin x \, dx + c$.Thus, $I = e^x \cos x + c$.

If an antiderivative of $f(x)$ is $e^x$ and an antiderivative of $g(x)$ is $\cos x$, then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx =$

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