$\displaystyle\int e^x\bigl(f(x) + f'(x)\bigr)\,dx$ equals:
- $f(x) f'(x) + C$
- $e^x f(x) + C$
- $e^x f'(x) + C$
- $e^x\bigl(f(x) - f'(x)\bigr) + C$
Show answer and solution
Correct answer: B. $e^x f(x) + C$
This is the standard 'product with exponential' identity: $\dfrac{d}{dx}\bigl(e^x f(x)\bigr) = e^x f(x) + e^x f'(x) = e^x(f(x) + f'(x))$
so by anti-differentiation the integral is $e^x f(x) + C$.
A useful trick whenever an integrand splits as $e^x$ times (something + its derivative).