Discussion
The value of the definite integral
\(\int_{0}^{2} \frac{1}{3^{x} + 3}\,dx\)
is
Answer: B
The integral equals \(\frac{1}{3}\).
Split the integrand so that one part has its numerator equal to the derivative of its denominator (up to a constant):
\[ \begin{aligned} \frac{1}{3^{x} + 3} &= \frac{1}{3}\cdot\frac{3}{3^{x} + 3} \\ &= \frac{1}{3}\left(1 - \frac{3^{x}}{3^{x} + 3}\right) \end{aligned} \]
Since \(\frac{d}{dx}(3^{x} + 3) = 3^{x}\ln 3\), the second part integrates to a logarithm:
\[ \begin{aligned} \int_{0}^{2} \frac{3^{x}}{3^{x} + 3}\,dx \\ &= \frac{1}{\ln 3}\Big[\ln(3^{x} + 3)\Big]_{0}^{2} \\ &= \frac{\ln 12 - \ln 4}{\ln 3} = 1 \end{aligned} \]
So the integral is \(\frac{1}{3}(2 - 1) = \frac{1}{3}\).
The logarithm cancels exactly because \(\ln 12 - \ln 4 = \ln 3\), so the options containing \(\log_{e} 3\) keep a factor that disappears. \(\frac{1}{2}\) is too large: the integrand never exceeds \(\frac{1}{4}\) on \([0, 2]\), so the integral is below \(\frac{1}{2}\).
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