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If $\int \frac{e^x-1}{e^x+1} d x=f(x)+c$, then $f(x)$ is equal to
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Verified Answer
The correct answer is:
$2 \log \left(e^x+1\right)-x$
We have, $\int \frac{e^x-1}{e^x+1} d x$
$=\int\left(\frac{2 e^x}{e^x+1}-1\right) d x$
$=2 \log \left(e^x+1\right)-x+c$
On comparing with $f(x)+c$, we get $f(x)=2 \log \left(e^x+1\right)-x$
$=\int\left(\frac{2 e^x}{e^x+1}-1\right) d x$
$=2 \log \left(e^x+1\right)-x+c$
On comparing with $f(x)+c$, we get $f(x)=2 \log \left(e^x+1\right)-x$
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