Mathematics

$$\displaystyle\int \dfrac{1}{1+e^x}dx$$ is equal to?


ANSWER

$$log_e\left(\dfrac{e^x}{e^x+1}\right)+c$$


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

$$=-ln|1+e^{-x}|+c$$

$$=-ln\left|\dfrac{e^x+1}{e^x}\right|+c$$

$$=ln\left|\dfrac{e^x}{e^x+1}\right|+c$$.
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Single Correct Medium Published on 17th 09, 2020
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