Prove that $\int e^{g(x)} (g'(x) \cdot f(x) + f'(x)) dx = e^{g(x)} \cdot f(x)$.
Step-by-Step Solution
Key Concept: General
$I = \int e^{g(x)} (g'(x) f(x) + f'(x)) dx = \int e^{g(x)} g'(x) \cdot f(x) dx + \int e^{g(x)} f'(x) dx$. Integrate the first integral on the R.H.S. by parts taking $e^{g(x)} \cdot g'(x)$ as the second function, we get $I = e^{g(x)} f(x) - \int f'(x) e^{g(x)} dx + \int e^{g(x)} \cdot f'(x) dx = e^{g(x)} \cdot f(x)$
Correct Answer: $e^{g(x)} \cdot f(x)$