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If $\int e^{2 x} f^{\prime}(x) d x=g(x)$, then $\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right] d x=$
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The correct answer is:
$\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C$
We have,
$$
\begin{aligned}
& \int e^{2 x} f^{\prime}(x) d x=g(x) \\
& \text { Let } I=\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right) d x \\
& \left.=f(x) \int e^{2 x} d x-\int f^{\prime}(x) \int e^{2 x} d x\right) d x+\int e^{2 x} f^{\prime}(x) d x \\
& =\frac{f(x) e^{2 x}}{2}-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x+\int e^{2 x} f^{\prime}(x) d x \\
& =\frac{e^{2 x}}{2} f(x)-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x \\
& =\frac{1}{2}\left[e^{2 x} f(x)-\int e^{2 x} f^{\prime}(x) d x\right] \\
& =\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C
\end{aligned}
$$
$$
\begin{aligned}
& \int e^{2 x} f^{\prime}(x) d x=g(x) \\
& \text { Let } I=\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right) d x \\
& \left.=f(x) \int e^{2 x} d x-\int f^{\prime}(x) \int e^{2 x} d x\right) d x+\int e^{2 x} f^{\prime}(x) d x \\
& =\frac{f(x) e^{2 x}}{2}-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x+\int e^{2 x} f^{\prime}(x) d x \\
& =\frac{e^{2 x}}{2} f(x)-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x \\
& =\frac{1}{2}\left[e^{2 x} f(x)-\int e^{2 x} f^{\prime}(x) d x\right] \\
& =\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C
\end{aligned}
$$
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