Evaluate Definite Integral of Rational Exponential Expression

To evaluate the definite integral $I = \int_0^2 \frac{1}{3^x + 3} dx$, we factor out $3$ from the denominator to write $I = \frac{1}{3} \int_0^2 \frac{1}{3^{x-1} + 1} dx$.

By making the substitution $u = x - 1$, the integral transforms into symmetric limits from $-1$ to $1$: $I = \frac{1}{3} \int_{-1}^1 \frac{1}{3^u + 1} du$

Using the integral property $\int_{-a}^a \frac{1}{f^u + 1} du = a$ for any $f > 0$, we find that the integral evaluates directly to $1$.

Thus, $I = \frac{1}{3} \times 1 = \frac{1}{3}$, which corresponds to option B.