If $\int \frac{e^x-1}{e^x+1} d x=f(x)+C$, then $f(x)=$

$\log \left(e^{2 x}-1\right)+C$

$\log \left(e^{2 x}+1\right)+C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

\int \frac{e^x-1}{e^x+1} d x=f(x)+C

, then

f(x)=

View answer and explanation

Correct option: C

2 \log \left(e^x+1\right)-x+C

Option C is correct because the integral simplifies to 2log(e^x + 1) - x + C after algebraic manipulation and integration.

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