The value of int frac{e^x}{e^x + 1} ,dx is | vedclass

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(C) Let $I = \int \frac{e^x}{e^x + 1} \,dx$. Substitute $e^x + 1 = t$. Differentiating both sides with respect to $x$,we get $e^x \,dx = dt$. Substitu

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$\log(e^x + 1) + c$

Vedclass · Answer

(C) Let $I = \int \frac{e^x}{e^x + 1} \,dx$. Substitute $e^x + 1 = t$. Differentiating both sides with respect to $x$,we get $e^x \,dx = dt$. Substituting these into the integral,we get $I = \int \frac{1}{t} \,dt$. The integral of $\frac{1}{t}$ is $\log|t| + c$. Substituting back $t = e^x + 1$,we get $I = \log|e^x + 1| + c$. Since $e^x + 1$ is always positive for all real $x$,we can write $I = \log(e^x + 1) + c$.

The value of $\int \frac{e^x}{e^x + 1} \,dx$ is

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