Integral Calculus

Concept:Use the formula $\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C$ for $k \neq 0$.Explanation:Here $k = -3$, so $\int e^{-3x} \, dx = \frac{1}{-3} e^{-3x} + C = -\frac{1}{3} e^{-3x} + C$.None of the given options match this result exactly. Option A and B have $+\frac{1}{3}e^{-3x}$, which is missing the negative sign; Option D has $-3e^{-3x}$, which is 9 times too large; Option C has $+\frac{1}{3}e^{3x}$, which is the integral of $e^{3x}$.If the intended integral was $\int e^{3x} \, dx$, then $\frac{1}{3}e^{3x}+C$ is correct.Answer:$-\frac{1}{3}e^{-3x}+C$ (not listed). Assuming a misprint, the closest matching option is $\frac{1}{3}e^{3x}+C$.