Mathematics

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


SOLUTION
$$\displaystyle \int{\dfrac{1}{{e}^{x}+1}dx}$$
$$=\displaystyle \int{\dfrac{{e}^{-x}}{{e}^{-x}\left({e}^{x}+1\right)}dx}$$
$$=\displaystyle \int{\dfrac{{e}^{-x}}{1+{e}^{-x}}dx}$$
Let $$t={e}^{-x}+1\Rightarrow dt=-{e}^{-x}dx$$
$$=\displaystyle \int{\dfrac{-dt}{t}}$$
$$=-\ln{\left|t\right|}+c$$ where $$c$$ is the constant of integration.
$$=-\ln{\left|{e}^{-x}+1\right|}+c$$
$$=-\ln{\left|\dfrac{1}{{e}^{x}}+1\right|}+c$$
$$=-\left(\ln{\left|\dfrac{1+{e}^{x}}{{e}^{x}}\right|}\right)+c$$
$$=-\left(\ln{\left(1+{e}^{x}\right)}-\ln{{e}^{x}}\right)+c$$
$$=-\ln{\left(1+{e}^{x}\right)}+\ln{{e}^{x}}+c$$
$$=-\ln{\left(1+{e}^{x}\right)}+x+c$$
$$=x-\ln{\left(1+{e}^{x}\right)}+c$$

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Subjective Medium Published on 17th 09, 2020
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