Question

If \[ \frac{d}{dx}\bigl(F(x)\bigr)=\frac{1}{e^x+1}, \] then find \(F(x)\), given that the initial condition is \[ F(0)=\log_e\!\left(\frac{1}{2}\right). \]

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!

Answer 1

Concept: The fundamental theorem of calculus states that if the derivative of a function is known, the function itself can be retrieved by finding the indefinite integral: \[ F(x) = \int \frac{d}{dx}(F(x)) \, dx = \int \frac{1}{e^x + 1} \, dx \] To solve integrals involving \(e^x\), a common technique is to multiply the numerator and denominator by \(e^{-x}\) or substitute variables to create a form matching standard log rules: \( \int \frac{f'(x)}{f(x)} \, dx = \log_e|f(x)| \).

Step 1: Set up the indefinite integration problem.
Given the derivative function, we write: \[ F(x) = \int \frac{1}{e^x + 1} \, dx \]

Step 2: Perform algebraic manipulation on the integrand.
Multiply both the numerator and the denominator by \(e^{-x}\): \[ F(x) = \int \frac{1 \cdot e^{-x}}{(e^x + 1) \cdot e^{-x}} \, dx = \int \frac{e^{-x}}{e^x \cdot e^{-x} + 1 \cdot e^{-x}} \, dx \] \[ F(x) = \int \frac{e^{-x}}{1 + e^{-x}} \, dx \]

Step 3: Solve the integral using substitution method.
Let us choose a new variable \( u = 1 + e^{-x} \). Differentiating both sides with respect to \( x \): \[ \frac{du}{dx} = 0 - e^{-x} \quad \Rightarrow \quad du = -e^{-x} \, dx \quad \Rightarrow \quad e^{-x} \, dx = -du \] Substitute these expressions into our integral equation: \[ F(x) = \int \frac{-du}{u} = -\int \frac{1}{u} \, du \] Integrating this yields a natural logarithm function plus an integration constant \(C\): \[ F(x) = -\log_e|u| + C \] Substitute back the expression for \( u \): \[ F(x) = -\log_e(1 + e^{-x}) + C \quad \cdots (1) \] We can rewrite the term inside the log to check compatibility: \[ 1 + e^{-x} = 1 + \frac{1}{e^x} = \frac{e^x + 1}{e^x} \] Substituting this version back into the equation: \[ F(x) = -\log_e\left(\frac{e^x + 1}{e^x}\right) + C = -\left[\log_e(e^x + 1) - \log_e(e^x)\right] + C \] Since \(\log_e(e^x) = x\), this simplifies neatly to: \[ F(x) = x - \log_e(e^x + 1) + C \quad \cdots (2) \]

Step 4: Use the given initial boundary condition to solve for \(C\).
We are given that \( F(0) = \log_e\left(\frac{1}{2}\right) = -\log_e(2) \). Let us use form (2) and set \( x = 0 \): \[ F(0) = 0 - \log_e(e^0 + 1) + C \] Since \(e^0 = 1\): \[ -\log_e(2) = -\log_e(1 + 1) + C \] \[ -\log_e(2) = -\log_e(2) + C \quad \Rightarrow \quad C = 0 \]

Step 5: Final expression formulation.
Substitute \(C = 0\) back into Equation (2): \[ F(x) = x - \log_e(1 + e^x) \]