Mathematics

Evaluate $$\int {\frac {e^{2x}}{1+e^x}}dx$$


SOLUTION
$$\displaystyle l = \int \dfrac{e^{2x}}{1+e^{x}}dx $$
$$\displaystyle = \int \dfrac{e^{x}e^{x}}{1+e^{x}}dx $$ 
Let $$ (1+e^{x}) = t $$
$$ e^{x} = (t-1) $$
$$ e^{x}dx = dt $$
$$\displaystyle l = \int \dfrac{(t-1)}{t}dt $$
$$\displaystyle \int (1-\dfrac{1}{t})dt $$
$$ = (t-logt)+c $$
$$ = e^{x}-loge^{x}+c $$
$$ = e^{x}-x+c $$ 
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Subjective Medium Published on 17th 09, 2020
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