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

Concept: Integration with a substitution to handle the exponential denominator, then applying the initial condition.

We need to integrate $\frac{1}{e^x+1}$.

Rewrite the integrand by multiplying numerator and denominator by $e^{-x}$:

Let $u = 1 + e^{-x}$, so $du = -e^{-x}dx$, which gives $e^{-x}dx = -du$.

The integral becomes:

Apply the initial condition $F(0) = \log\frac{1}{2}$:

$-\log(1+e^0) + C = \log\frac{1}{2}$ …

The problem asks us to find an antiderivative of $\frac{1}{e^x+1}$. At first glance, this doesn't match any standard form. The key insight is to manipulate the integrand algebraically so that it becomes recognizable as the derivative of a logarithmic function.

The denominator $e^x + 1$ suggests we might want to work with $\log(e^x+1)$, but differentiating that gives $\frac{e^x}{e^x+1}$, not quite what we have. The trick is to multiply both numerator and denominator by $e^{-x}$, which transforms the expression without changing its value.

Finding the antiderivative:

1. Rewrite the integrand by multiplying by $\frac{e^{-x}}{e^{-x}}$:

2. Recognize the derivative pattern. Notice that if we let $u = 1 + e^{-x}$, then $\frac{du}{dx} = -e^{-x}$. This means:

3. Integrate both sides. Since $\frac{d}{dx}(F(x)) = \frac{1}{e^x+1} = -\frac{d}{dx}\left[\log(1+e^{-x})\right]$, we have:

where $C$ is the constant of integration.

4. Apply the initial condition $F(0) = \log \frac{1}{2}$: …