If int e^{2x} f^{prime}(x) dx = g(x),then int | vedclass

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(B) We are given that $\int e^{2x} f^{\prime}(x) dx = g(x)$.Let $I = \int (e^{2x} f(x) + e^{2x} f^{\prime}(x)) dx$.This can be split into two integral

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

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(B) We are given that $\int e^{2x} f^{\prime}(x) dx = g(x)$.Let $I = \int (e^{2x} f(x) + e^{2x} f^{\prime}(x)) dx$.This can be split into two integrals: $I = \int e^{2x} f(x) dx + \int e^{2x} f^{\prime}(x) dx$.Using integration by parts for the first integral $\int e^{2x} f(x) dx$ (taking $f(x)$ as the first function and $e^{2x}$ as the second function):$\int e^{2x} f(x) dx = f(x) \int e^{2x} dx - \int (f^{\prime}(x) \int e^{2x} dx) dx = \frac{1}{2} f(x) e^{2x} - \frac{1}{2} \int e^{2x} f^{\prime}(x) dx$.Substituting this back into the expression for $I$:$I = [\frac{1}{2} f(x) e^{2x} - \frac{1}{2} \int e^{2x} f^{\prime}(x) dx] + \int e^{2x} f^{\prime}(x) dx$.$I = \frac{1}{2} e^{2x} f(x) + \frac{1}{2} \int e^{2x} f^{\prime}(x) dx$.Since $\int e^{2x} f^{\prime}(x) dx = g(x)$,we have:$I = \frac{1}{2} e^{2x} f(x) + \frac{1}{2} g(x) + C = \frac{1}{2} [e^{2x} f(x) + g(x)] + C$.

If $\int e^{2x} f^{\prime}(x) dx = g(x)$,then $\int (e^{2x} f(x) + e^{2x} f^{\prime}(x)) dx =$

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