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: Integrate the given derivative
We are given that $\dfrac{d}{dx}(F(x)) = \dfrac{1}{e^x + 1}$. To find $F(x)$, we need to integrate the right-hand side: \[ F(x) = \int \dfrac{1}{e^x + 1} \, dx. \] Let us perform the integration. Consider the substitution $u = e^x + 1$, so that $du = e^x dx$. Therefore: \[ F(x) = \int \dfrac{1}{u} \, du = \log|u| + C = \log(e^x + 1) + C. \]

Step 2: Use the initial condition
We are given the condition $F(0) = \log\left(\dfrac{1}{2}\right)$. Substituting $x = 0$ into the expression for $F(x)$: \[ F(0) = \log(e^0 + 1) + C = \log(1 + 1) + C = \log(2) + C. \] We are given that $F(0) = \log\left(\dfrac{1}{2}\right)$, so: \[ \log(2) + C = \log\left(\dfrac{1}{2}\right). \] Since $\log\left(\dfrac{1}{2}\right) = -\log(2)$, we have: \[ \log(2) + C = -\log(2), \] which implies: \[ C = -2\log(2). \]

Step 3: Final expression for $F(x)$
Now that we have determined the value of $C$, the function $F(x)$ is: \[ F(x) = \log(e^x + 1) - 2\log(2). \] This is the required function $F(x)$.

Final Answer:
\[ F(x) = \log(e^x + 1) - 2\log(2). \]