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# Integrate : $\int \large\frac {e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$$dx $\begin {array} {1 1} (a)\;\log (e^{2x}-e^{-2x})+c \\ (b)\;\log (e^{2x}+e^{-2x})+c \\ (c)\;\log (e^{2x}-e^{-2x})^{1/2}+c \\ (d)\;None \end {array}$ Can you answer this question? ## 1 Answer 0 votes e^{2x}-e^{-2x}=t => 2[e^{2x}+e^{-2x}]dx=dt => \int \large\frac{dt}{2t} \large\frac{1}{2}$$\log (e^{2x}-e^{-2x})+c$
Hence c is the correct answer.