​​​​​​​Đáp án:

\( 2^{f(x)} + f(x) = x + 1 \Rightarrow f'(x) 2^{f(x)} + f'(x)f(x) = x f'(x) + f'(x) \)

\( \Rightarrow \int_{0}^{2} f'(x) \cdot 2^{f(x)} dx + \int_{0}^{2} f'(x) \cdot f(x) dx = \int_{0}^{2} x f'(x) dx + \int_{0}^{2} f'(x) dx \)

\(  \frac{2^{f(x)}}{\ln 2} \Big|_{0}^{2} + \frac{(f(x))^2}{2} \Big|_{0}^{2} -  f(x) \Big|_{0}^{2}  = \int_{0}^{2} f(x) dx \)

Đặt: \( \begin{cases} u = x \\ dv = f'(x) dx \end{cases} \Rightarrow \begin{cases} du=dx \\ v= f(x) \end{cases} \)

\( \int_{0}^{2} x f'(x) dx = x f(x) \Big|_{0}^{2} - \int_{0}^{2} f(x) dx \)

\( \Rightarrow \int_{0}^{2} f(x) dx = x f(x) \Big|_{0}^{2} - \frac{2^{f(x)}}{\ln 2} \Big|_{0}^{2} - \frac{(f(x))^2}{2} \Big|_{0}^{2} + f(x) \Big|_{0}^{2} \)

+ \( 2^{f(x)} + f(x) = x + 1, \forall x \in \mathbb{R} \)

\( x = 0: 2^{f(0)} + f(0) = 1 \Leftrightarrow f(0) = 0 \)

\( x = 2: 2^{f(2)} + f(2) = 3 \Leftrightarrow f(2) = 1 \)

\( \Rightarrow \int_{0}^{2} f(x) dx = 2 - \frac{1}{\ln 2} (2 - 1) - \frac{1}{2} + 1 \)

\( = \frac{5}{2} - \frac{1}{\ln 2} \Rightarrow \begin{cases} a = 5\\ b = -1 \end{cases} \Rightarrow a + b = 4 \Rightarrow \boxed{A} \)