If int frac{e^x-1}{e^x+1} d x=f(x)+c,then f(x | vedclass

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(C) We have the integral $I = \int \frac{e^x-1}{e^x+1} d x$.We can rewrite the integrand as $\frac{e^x-1}{e^x+1} = \frac{2e^x - (e^x+1)}{e^x+1} = \fra

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

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(C) We have the integral $I = \int \frac{e^x-1}{e^x+1} d x$.We can rewrite the integrand as $\frac{e^x-1}{e^x+1} = \frac{2e^x - (e^x+1)}{e^x+1} = \frac{2e^x}{e^x+1} - 1$.Now,integrating term by term,we get $I = \int \frac{2e^x}{e^x+1} d x - \int 1 d x$.For the first part,let $u = e^x+1$,then $du = e^x d x$.Thus,$\int \frac{2e^x}{e^x+1} d x = 2 \log |e^x+1|$.Therefore,$I = 2 \log (e^x+1) - x + c$.Comparing this with $f(x)+c$,we find $f(x) = 2 \log (e^x+1) - x$.

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

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