4.0 Integral of the type $\int {{e^x}\left\{ {f(x) + f'(x)} \right\}dx} $

The solution of the integral of the type $\int {{e^x}\left\{ {f(x) + f'(x)} \right\}dx} $ is ${e^x}f(x) + C$

Proof: $$\begin{equation} \begin{aligned} I = \int {{e^x}\left\{ {f(x) + f'(x)} \right\}dx} \\ I = \int {{e^x}} f(x)dx + \int {{e^x}} f'(x)dx \quad ...(i) \\ I = {I_1} + {I_2} \\\end{aligned} \end{equation} $$

Now, take ${I_1}$ and apply integration by parts, we get, $$\begin{equation} \begin{aligned} {I_1} = \int {{e^x}} f(x)dx \\ {I_1} = f(x).{e^x} - \int {f'(x)} .{e^x}dx + C \quad ...(ii)\\\end{aligned} \end{equation} $$

Therefore from equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} I = f(x).{e^x} - \int {f'(x)} .{e^x}dx + \int {f'(x)} .{e^x}dx + C \\ I = f(x).{e^x} + C \\\end{aligned} \end{equation} $$