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If $\int e^x\left(f(x)-f^{\prime}(x)\right)=g(x)+C$, then $\int e^x f^{\prime}(x) d x=$
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Verified Answer
The correct answer is:
$\frac{1}{2}\left[e^x f(x)-g(x)\right]+C$
We are given that
$$
\begin{aligned}
& \int e^x\left[f(x)-f^{\prime}(x)\right] d x=g(x)+c \\
& \Rightarrow \int e^x f(x)-\int e^x f^{\prime}(x)=g(x)+c \\
& \Rightarrow \int e^x \cdot f(x) d x=\int e^x f^{\prime}(x)-g(x)+c \\
& \text { Now } \int e^x \cdot f^{\prime}(x) d x= \\
& f(x) \cdot e^x-\int f^{\prime}(x) \cdot e^x d x-g(x)+c \\
& \Rightarrow 2 \int e^x \cdot f^{\prime}(x) d x=e^x \cdot f(x)-g(x)+c \\
& \Rightarrow \int e^x \cdot f^{\prime}(x) d x=\frac{1}{2}\left[e^x \cdot f(x)-g(x)\right]+c
\end{aligned}
$$
$$
\begin{aligned}
& \int e^x\left[f(x)-f^{\prime}(x)\right] d x=g(x)+c \\
& \Rightarrow \int e^x f(x)-\int e^x f^{\prime}(x)=g(x)+c \\
& \Rightarrow \int e^x \cdot f(x) d x=\int e^x f^{\prime}(x)-g(x)+c \\
& \text { Now } \int e^x \cdot f^{\prime}(x) d x= \\
& f(x) \cdot e^x-\int f^{\prime}(x) \cdot e^x d x-g(x)+c \\
& \Rightarrow 2 \int e^x \cdot f^{\prime}(x) d x=e^x \cdot f(x)-g(x)+c \\
& \Rightarrow \int e^x \cdot f^{\prime}(x) d x=\frac{1}{2}\left[e^x \cdot f(x)-g(x)\right]+c
\end{aligned}
$$
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