To determine whether the improper integral
$\int_{0}^{\infty} \frac{e^x}{1 + e^{2x}} \, dx$ converges and evaluate its value, we proceed as follows:

Step 1: Simplify the integrand

Start with the integrand: $f(x) = \frac{e^x}{1 + e^{2x}}.$

We can factor out $e^x$ in the denominator as follows: $1 + e^{2x} = e^{2x} \left( e^{-2x} + 1 \right).$ Thus: $\frac{e^x}{1 + e^{2x}} = \frac{e^x}{e^{2x} (e^{-2x} + 1)} = \frac{e^x}{e^{2x}} \cdot \frac{1}{1 + e^{-2x}}.$

Simplify the powers of $e$: $\frac{e^x}{e^{2x}} = e^{-x}.$ Therefore: $f(x) = \frac{e^{-x}}{1 + e^{-2x}}.$

Step 2: Substitution to simplify the integral

To make progress, use the substitution: $u = e^{-x}, \quad \text{so} \quad du = -e^{-x} dx \quad \Rightarrow \quad dx = -\frac{du}{u}.$

Now update the limits of integration:

• When $x = 0$, $u = e^{-0} = 1$,
• When $x \to \infty$, $u \to 0$.

Substituting into the integral: $\int_{0}^{\infty} \frac{e^x}{1 + e^{2x}} \, dx = \int_{1}^{0} \frac{u}{1 + u^2} \cdot \left( -\frac{du}{u} \right).$

Simplify: $\int_{0}^{\infty} \frac{e^x}{1 + e^{2x}} \, dx = \int_{0}^{1} \frac{1}{1 + u^2} \, du.$

Step 3: Evaluate the integral

The integral of $\frac{1}{1 + u^2}$ is the standard result: $\int \frac{1}{1 + u^2} \, du = \arctan(u).$

Thus: $\int_{0}^{1} \frac{1}{1 + u^2} \, du = \arctan(1) - \arctan(0).$

Using the known values of the arctangent function: $\arctan(1) = \frac{\pi}{4}, \quad \arctan(0) = 0.$

Therefore: $\int_{0}^{\infty} \frac{e^x}{1 + e^{2x}} \, dx = \frac{\pi}{4}.$

Final Answer:

The integral converges and evaluates to: $\boxed{\frac{\pi}{4}}$

Thus, the correct choice is (a).

Do you want further clarification on any step? Let me know!

Related Questions:

1. What makes an improper integral converge or diverge?
2. How do you use substitution to evaluate improper integrals?
3. What is the standard integral for $\frac{1}{1 + x^2}$ and how is it derived?
4. Why is $e^{-x}$ a useful substitution in this problem?
5. How can improper integrals be applied in real-world problems?

Tip:

For improper integrals, always verify whether the integral converges before evaluating it. Substitution and standard integral forms often simplify the process!