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 Q19  Integrate the functions \frac{e ^{2x}-1}{e ^{2x}+1} 

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Given function  \frac{e ^{2x}-1}{e ^{2x}+1},

Simplifying it by dividing both numerator and denominator by e^x, we obtain

\frac{\frac{e^{2x}-1}{e^x}}{\frac{e^{2x}+1}{e^x}} = \frac{e^x-e^{-x}}{e^x+e^{-x}}

Assume the e^{x}+e^{-x} =t

 \therefore (e^x-e^{-x})dx =dt

\implies \int \frac{e^{2x}-1}{e^{2x}+1}dx = \int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx

= \int \frac{dt}{t}

= \log |t| +C

Now, back substituting the value of t,

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

Posted by

Divya Prakash Singh

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