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This integral can be solved using u-substitution.Let ( u = e^{2x} - e^{-2x} ).Then, find ( du ) by differentiating ( u ) with respect to ( x ):( \frac{du}{dx} = \frac{d}{dx}(e^{2x}) - \frac{d}{dx}(e^{-2x}) = 2e^{2x} - (-2e^{-2x}) = 2e^{2x} + 2e^{-2x} )So, ( du = (2e^{2x} + 2e^{-2x}) dx = 2(e^{2x} + e^{-2x}) dx ).Rearrange to find ( (e^{2x} + e^{-2x}) dx ): ( (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 )The integral of ( \frac{1}{u} ) is ( \ln|u| ):( = \frac{1}{2} \ln|u| + C )Substitute back ( u = e^{2x} - e^{-2x} ):( = \frac{1}{2} \ln|e^{2x} - e^{-2x}| + C ).This matches option C.