Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 363: 77

$\frac{ln(3^{2x}+1)}{2ln(3)}+C$

let u=$3^{2x}+1$ $2ln(3)3^{2x}$ $dx=du$ $dx=\frac{1}{2ln(3)3^{2x}}du$ $\int\frac{3^{2x}}{1+3^{2x}}dx$ $=\frac{1}{2ln(3)}\int \frac{1}{u}du$ $=\frac{ln(u)}{2ln(3)}+C$ $=\frac{ln(3^{2x}+1)}{2ln(3)}+C$