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$\int 2^{x}\left[f^{\prime}(x)+f(x) \log 2\right] d x$ is equal to
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The correct answer is:
$2^{x} f(x)+C$
Let $I=\int 2^{n}\left[f^{\prime}(x)+f(x) \log 2\right] d x$
$\begin{array}{ll}\text { Consider } & g(x)=2^{x} f(x) \\ \Rightarrow & g^{\prime}(x)=2^{x} f^{\prime}(x)+2^{x} f(x) \log 2 \\ \Rightarrow & g^{\prime}(x)=2^{x}\left[f^{\prime}(x)+f(x) \log 2\right] \\ \therefore \quad & \quad I=\int g^{\prime}(x) d x=g(x)+C=2^{x} f(x)+C\end{array}$
$\begin{array}{ll}\text { Consider } & g(x)=2^{x} f(x) \\ \Rightarrow & g^{\prime}(x)=2^{x} f^{\prime}(x)+2^{x} f(x) \log 2 \\ \Rightarrow & g^{\prime}(x)=2^{x}\left[f^{\prime}(x)+f(x) \log 2\right] \\ \therefore \quad & \quad I=\int g^{\prime}(x) d x=g(x)+C=2^{x} f(x)+C\end{array}$
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