If int e^x left(f(x) - f^{prime}(x)right) dx | vedclass

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(A) Given that $\int e^x \left[f(x) - f^{\prime}(x)\right] dx = g(x) + C$. Expanding the integral,we have $\int e^x f(x) dx - \int e^x f^{\prime}(x) d

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$\frac{1}{2} \left[e^x f(x) - g(x)\right] + C$

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(A) Given that $\int e^x \left[f(x) - f^{\prime}(x)\right] dx = g(x) + C$. Expanding the integral,we have $\int e^x f(x) dx - \int e^x f^{\prime}(x) dx = g(x) + C$. Rearranging,we get $\int e^x f(x) dx = \int e^x f^{\prime}(x) dx + g(x) + C$. Using integration by parts on $\int e^x f(x) dx$: $\int e^x f(x) dx = f(x) e^x - \int e^x f^{\prime}(x) dx$. Substituting this into our rearranged equation: $f(x) e^x - \int e^x f^{\prime}(x) dx = \int e^x f^{\prime}(x) dx + g(x) + C$. $f(x) e^x - g(x) = 2 \int e^x f^{\prime}(x) dx + C$. Therefore,$\int e^x f^{\prime}(x) dx = \frac{1}{2} \left[e^x f(x) - g(x)\right] + C$.

If $\int e^x \left(f(x) - f^{\prime}(x)\right) dx = g(x) + C$,then $\int e^x f^{\prime}(x) dx =$

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