If
\[
\frac{d}{dx}\bigl(F(x)\bigr)=\frac{1}{e^x+1},
\]
then find \(F(x)\), given that the initial condition is
\[
F(0)=\log_e\!\left(\frac{1}{2}\right).
\]
Show Hint
An alternative trick for this integral is adding and subtracting \(e^x\) in the numerator:
\[
\int \frac{1}{e^x+1}dx = \int \frac{(1+e^x) - e^x}{e^x+1}dx = \int 1 dx - \int \frac{e^x}{e^x+1}dx = x - \log_e(e^x+1) + C
\]
This avoids negative exponents entirely and directly yields the final functional form in seconds!