Question Detail

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