Question Detail

This integral can be solved using u-substitution.Let the denominator be $u$: $u = e^{2x} - e^{-2x}$.Now, find the differential $du$:$du = \frac{d}{dx}(e^{2x} - e^{-2x}) dx = (2e^{2x} - (-2e^{-2x})) dx = (2e^{2x} + 2e^{-2x}) dx = 2(e^{2x} + e^{-2x}) dx$.From this, we can see that $(e^{2x} + e^{-2x}) dx = \frac{1}{2} du$.Substitute $u$ and $du$ into the integral:$\int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du$.Integrate with respect to $u$:$= \frac{1}{2} \ln|u| + C$.Substitute back $u = e^{2x} - e^{-2x}$:$= \frac{1}{2} \ln|e^{2x} - e^{-2x}| + C$.