Question

If $\dfrac{d}{dx}(F(x))=\dfrac{1}{e^x+1}$, then find $F(x)$ given that $F(0)=\log\left(\dfrac{1}{2}\right)$.

Hint:

When integrating expressions involving \(e^x\) in denominators, multiplying numerator and denominator by \(e^{-x}\) often simplifies the integral and allows easy substitution.

Answer 1

Step 1: Write the given differential relation.
We are given that \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1} \] To obtain \(F(x)\), we integrate both sides with respect to \(x\).
Step 2: Integrate the expression.
\[ F(x)=\int \frac{1}{e^x+1}\,dx \] Rewrite the integrand by multiplying numerator and denominator by \(e^{-x}\): \[ \frac{1}{e^x+1}=\frac{e^{-x}}{1+e^{-x}} \] Let \[ u=1+e^{-x} \] Then \[ \frac{du}{dx}=-e^{-x} \] \[ du=-e^{-x}\,dx \] Thus \[ e^{-x}dx=-du \] Substitute in the integral: \[ \int \frac{e^{-x}}{1+e^{-x}}dx = \int \frac{-du}{u} \] \[ =-\ln|u|+C \] Substituting \(u=1+e^{-x}\): \[ F(x)=-\ln(1+e^{-x})+C \] Step 3: Use the initial condition.
Given \[ F(0)=\log\left(\frac12\right) \] Substitute \(x=0\): \[ F(0)=-\ln(1+e^{0})+C \] \[ =-\ln(2)+C \] But \[ F(0)=\ln\left(\frac12\right)=-\ln 2 \] Thus \[ -\ln2=-\ln2+C \] \[ C=0 \] Step 4: Write the final function.
Therefore \[ F(x)=-\ln(1+e^{-x}) \] Step 5: Equivalent logarithmic form.
Using logarithmic properties, \[ -\ln(1+e^{-x})=\ln\left(\frac{1}{1+e^{-x}}\right) \] Hence the required function can also be written as \[ F(x)=\ln\left(\frac{1}{1+e^{-x}}\right) \] Final Answer:
\[ \boxed{F(x)=-\ln(1+e^{-x})} \]