1. The problem states that $$\int (f(x))^n \cdot g(x) \, dx = \frac{1}{n+1} [f(x)]^{n+1} + c$$. 2. To find $$g(x)$$, differentiate both sides with respect to $$x$$ using the Fundamental Theorem of Calculus: $$\frac{d}{dx} \left( \int (f(x))^n \cdot g(x) \, dx \right) = \frac{d}{dx} \left( \frac{1}{n+1} [f(x)]^{n+1} + c \right)$$ 3. The left side simplifies to: $$(f(x))^n \cdot g(x)$$ 4. The right side differentiates as: $$\frac{1}{n+1} \cdot (n+1) [f(x)]^n \cdot f'(x) = [f(x)]^n \cdot f'(x)$$ 5. Equate both sides: $$(f(x))^n \cdot g(x) = (f(x))^n \cdot f'(x)$$ 6. Since $$ (f(x))^n \neq 0 $$, divide both sides by $$ (f(x))^n $$: $$g(x) = f'(x)$$ 7. Therefore, the correct answer is (b) $$f'(x)$$. Final answer: $$g(x) = f'(x)$$.