\( \int(x+1)e^x\,dx \) is equal to
[ \( C \) is an arbitrary constant]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( xe^x+C \)
The integrand has the form \( e^x\left[f(x)+f'(x)\right] \), and
\( \displaystyle \int e^x\left[f(x)+f'(x)\right]dx=e^x f(x)+C \)
With f(x) = x, f′(x) = 1, so \( x+1=f(x)+f'(x) \):
\( \displaystyle \int(x+1)e^x\,dx=xe^x+C \)
Shortcut: Check by differentiating Option (3): \( \frac{d}{dx}(xe^x)=xe^x+e^x=(x+1)e^x \).