Mathematics

solve $$\displaystyle\int\dfrac { e^{ 2x }-{ e }^{ -2x } }{ { e }^{ 2x }+{ e }^{ -2c } }dx $$


SOLUTION
$$\displaystyle\int\dfrac { e^{ 2x }-{ e }^{ -2x } }{ { e }^{ 2x }+{ e }^{ -2c } }dx $$

Put ,  $$e^{2x} + e^{-2x} = t $$
$$ \implies (e^{2x}(2) + e^{-2x}(- 2) ) dx = dt$$

$$\implies ( e^{2x} - e^{-2x}) dx = \dfrac{1}{2} dt$$

Now, $$\displaystyle\int \dfrac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx = \int \dfrac{1}{2t} dt$$

$$= \dfrac{1}{2} \log {t} + c$$

$$= \dfrac{1}{2} \log (e^{2x} + e^{-2x}) + c$$
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Subjective Medium Published on 17th 09, 2020
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