Mathematics

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


ANSWER

$$2 \, \log (e^x + 1) - x$$


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

let $$u = e^x ; du = e^x dx $$

$$=\int \dfrac{u-1}{u(u+1)}du$$

$$\int  \dfrac {2u-(u+1)} {u(u+1)} du$$

$$=2\int  \dfrac{1}{u+1}du -\int \dfrac{1}{u} du$$

$$=2\ln (u+1)-\ln (u)+C$$

$$=2 \ln (e^x+1)-x+C$$
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Single Correct Medium Published on 17th 09, 2020
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