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

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(C) We have the integral $I = \int \frac{e^x - 1}{e^x + 1} dx$. To simplify the integrand, we can rewrite the numerator as $(e^x + 1) - 2$. Thus, $I =

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

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(C) We have the integral $I = \int \frac{e^x - 1}{e^x + 1} dx$. To simplify the integrand, we can rewrite the numerator as $(e^x + 1) - 2$. Thus, $I = \int \frac{(e^x + 1) - 2}{e^x + 1} dx$. $I = \int \left( \frac{e^x + 1}{e^x + 1} - \frac{2}{e^x + 1} \right) dx = \int \left( 1 - \frac{2}{e^x + 1} \right) dx$. Alternatively, multiply the numerator and denominator by $e^{-x}$: $I = \int \frac{1 - e^{-x}}{1 + e^{-x}} dx$. Let $u = 1 + e^{-x}$, then $du = -e^{-x} dx$, so $dx = -du / (u - 1)$. Using the first method: $I = \int 1 dx - 2 \int \frac{1}{e^x + 1} dx$. For $\int \frac{1}{e^x + 1} dx$, multiply by $e^{-x}/e^{-x}$: $\int \frac{e^{-x}}{1 + e^{-x}} dx$. Let $v = 1 + e^{-x}$, then $dv = -e^{-x} dx$. So, $\int \frac{1}{e^x + 1} dx = -\int \frac{1}{v} dv = -\log |1 + e^{-x}| = -\log |(e^x + 1)/e^x| = -\log (e^x + 1) + x$. Thus, $I = x - 2(-\log (e^x + 1) + x) + c = x + 2 \log (e^x + 1) - 2x + c = 2 \log (e^x + 1) - x + c$. Comparing with $f(x) + c$, we get $f(x) = 2 \log (e^x + 1) - x$.

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

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