Mathematics

# Integrate : $\dfrac {{e}^{2x}-1}{{e}^{2x}+1}$

##### SOLUTION
$\displaystyle \int \dfrac{e^{2x}-1}{e^{2x}+1} dx = \int \dfrac{e^x - \dfrac{1}{e^x}}{e^x + \dfrac{1}{e^x}} dx \int \dfrac{e^x - e^{-x}}{e^x + e^2} dx$

$\displaystyle e^x + e^{-x} = t \Rightarrow dx. [e^x - e^{-x}] = dt$

$\displaystyle \int \dfrac{dt}{t} = log | t | + c = log (e^x + e^{-x}) +c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
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