Explanation

(c)
Given, $\int_{}^{}{e^{x}\left( f(x) - f^{'}(x) \right)dx = \phi(x)}$
$\Rightarrow \ e^{x}\left( f(x) - f^{'}(x) \right) = \phi'(x)$
Let $I = \int_{}^{}{e^{x}f(x)dx}$
$= f(x)e^{x} - \int_{}^{}{f^{'}(x)e^{x}dx}$
$= f(x)e^{x} - \left\lbrack \ \int_{}^{}{e^{x}f(x)dx - \int_{}^{}\phi^{'}}(x)dx\ \right\rbrack$
$\Rightarrow 2I = f(x)e^{x} + \phi(x)$
$\therefore I = \frac{1}{2}\left\lbrack \ e^{x}f(x) + \phi(x) \right\rbrack$