Question

find integration of e2x-epower -2x / e2x +epower-2x dx​

Answer 1

SOLUTION

TO INTEGRATE

$\displaystyle \int\limits_{}^{} \frac{ {e}^{2x} - {e}^{ - 2x} }{{e}^{2x} + {e}^{ - 2x} } \: \, dx$

EVALUATION

Here we have to find

$\displaystyle \int\limits_{}^{} \frac{ {e}^{2x} - {e}^{ - 2x} }{{e}^{2x} + {e}^{ - 2x} } \: \, dx$

Let

$z = {e}^{2x} + {e}^{ - 2x}$

Differentiating both sides with respect to x we get

$\displaystyle \frac{dz}{dx} = 2 {e}^{2x} - 2 {e}^{ - 2x}$

$\implies \: \displaystyle \frac{dz}{2} = ( {e}^{2x} - {e}^{ - 2x} )dx$

$\displaystyle \int\limits_{}^{} \frac{ {e}^{2x} - {e}^{ - 2x} }{{e}^{2x} + {e}^{ - 2x} } \: \, dx$

$= \displaystyle \frac{1}{2} \int\limits_{}^{} \frac{ dz }{z}$

$= \displaystyle \frac{1}{2} \log |z| +c$

$= \displaystyle \frac{1}{2} \log | {e}^{2x} + {e}^{ - 2x} | +c$

Where C is integration constant

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