Let $\int e^x\left\{f(x)-f'(x)\right\} d x=\phi(x)$. Then, $\int e^x f(x) d x$ is equal to

$\frac{1}{2}\left\{\phi(x)+e^x f(x)\right\}$

We know that

$\int e^x\left\{f(x)+f'(x)\right\} d x=e^x f(x)$

It is given that

$\int e^x\left\{f(x)-f'(x)\right\} d x=\phi(x)$

Adding these two, we get

$2 \int e^x f(x) d x=\phi(x)+e^x f(x)$

$\int e^x f(x) d x=\frac{1}{2}\left\{\phi(x)+e^x f(x)\right\}$