Mathematics

If $$\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx =f(x) + c$$ then $$f(x) =$$


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

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

$$=\displaystyle \int \dfrac{e^x}{e^x + 1}dx $$$$-\displaystyle \int \dfrac{ e^{-x}}{e^{-x} + 1}dx $$ [Multiplying the numerator and denominator of the second integral by $$e^{-x}$$]

$$=\displaystyle \int \dfrac{d(e^x+1)}{e^x + 1}dx $$$$+\displaystyle \int \dfrac{d( e^{-x}+1)}{e^{-x} + 1}dx $$
$$=\log(1+e^x)+\log(1+e^{-x})+c$$ [$$c$$ being integrating constant]
$$=\log(1+e^x)+\log(1+e^{x})-x+c$$
$$=2\log(1+e^x)-x+c$$
Comparing this with the given integral we get,
$$f(x)=2\log(1+e^{x})-x$$
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Subjective Medium Published on 17th 09, 2020
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