Discussion
\(\int_{0}^{1} \dfrac{3^{2x}}{3^{2x} + 1}\,dx\) =
Answer: A
The integral is \(\frac{log_{e}5}{2log_{e}3}\).
Put \(u = 3^{2x} + 1\), so \(du = 2\ln 3 \cdot 3^{2x}dx\):
\[ \begin{aligned} \int_{0}^{1}\frac{3^{2x}}{3^{2x} + 1}dx &= \frac{1}{2\ln 3}\left[\ln(3^{2x} + 1)\right]_{0}^{1} \\ &= \frac{\ln 10 - \ln 2}{2\ln 3} = \frac{\ln 5}{2\ln 3} \end{aligned} \]
The other denominators (3 or 9 times \(\ln 3\)) come from differentiating \(3^{2x}\) wrongly, for example as \(3^{2x}\ln 9\) and then again multiplying by 2 or 3.
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