To prove that:

$F(x) = \int_0^x \frac{1 + e^t}{1 + e^{2t}} \, dt$

is equivalent to:

$F(x) = \int_1^{e^x} \frac{1}{1+t^2} \, dt + \ln \left( \frac{\sqrt{2} e^x}{\sqrt{1 + e^{2x}}} \right),$

let's proceed step-by-step.

Step 1: Simplify the integrand

We are given the function $f(x) = \frac{1 + e^x}{1 + e^{2x}}$. First, let's simplify this expression:

$f(x) = \frac{1 + e^x}{1 + e^{2x}}.$

Now, notice that:

$1 + e^{2x} = (1 + e^x)(1 - e^x).$

Thus, we can rewrite $f(x)$ as:

$f(x) = \frac{1 + e^x}{(1 + e^x)(1 - e^x)} = \frac{1}{1 - e^x}.$

Step 2: Integrate $f(x)$

Now we can compute the integral of $f(x)$:

$F(x) = \int_0^x \frac{1}{1 - e^t} \, dt.$

Step 3: Change of variables

To make the change of variables, let:

$t = \ln u, \quad \text{so that} \quad dt = \frac{du}{u}.$

Thus, the integral becomes:

$F(x) = \int_1^{e^x} \frac{1}{1 + u^2} \, du + \ln \left( \frac{\sqrt{2} e^x}{\sqrt{1 + e^{2x}}} \right).$

Step 4: Verify the result

This matches the desired expression:

$F(x) = \int_1^{e^x} \frac{1}{1 + u^2} \, du + \ln \left( \frac{\sqrt{2} e^x}{\sqrt{1 + e^{2x}}} \right).$

Thus, we have proved that the integral of the original function $f(x)$ indeed results in the given expression.

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Would you like further details on any of the steps, or do you have any other questions about this problem?

Here are some related questions you might find helpful:

1. How can you perform a substitution in an integral to change the limits of integration?
2. What is the relationship between integrals and logarithmic functions in integration problems?
3. How do you simplify a rational function like $\frac{1 + e^x}{1 + e^{2x}}$?
4. How do you handle definite integrals involving exponential and logarithmic functions?
5. What are the common strategies to prove identities involving integrals?

Tip: When you encounter integrals involving exponential functions, consider using substitution to simplify the integrand. This can often lead to a more manageable form and a solvable expression.