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 Q20   Integrate the functions \frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}

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Given function  \frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }},

Assume the e^{2x}+e^{-2x} =t

 \therefore (2e^{2x}-2e^{-2x})dx =dt

\implies \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx = \int \frac{dt}{2t}

= \frac{1}{2}\int \frac{1}{t}dt

= \frac{1}{2}\log|t| +C

Now, back substituting the value of t,

= \frac{1}{2}\log|e^{2x}+e^{-2x}| +C , where C is any constant value.

Posted by

Divya Prakash Singh

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