The given integral is:

$\int \frac{e^{\tan^{-1}x}}{1 + x^2} \, dx$

Step-by-Step Solution:

1. Observation of the denominator: The denominator $1 + x^2$ is the derivative of the inverse tangent function ($\tan^{-1}x$):

   $\frac{d}{dx} \left( \tan^{-1}x \right) = \frac{1}{1 + x^2}.$

2. Substitution: Let: $u = \tan^{-1}x.$ Then the derivative of $u$ is: $\frac{du}{dx} = \frac{1}{1 + x^2} \quad \implies \quad du = \frac{dx}{1 + x^2}.$

3. Rewriting the integral: Substituting $u = \tan^{-1}x$ and $du = \frac{dx}{1 + x^2}$, the integral becomes: $\int \frac{e^{\tan^{-1}x}}{1 + x^2} \, dx = \int e^u \, du.$

4. Integration: The integral of $e^u$ is straightforward: $\int e^u \, du = e^u + C,$ where $C$ is the constant of integration.

5. Substitute back $u = \tan^{-1}x$: Replacing $u$ with $\tan^{-1}x$, we get: $e^u = e^{\tan^{-1}x}.$

6. Final Answer: $\int \frac{e^{\tan^{-1}x}}{1 + x^2} \, dx = e^{\tan^{-1}x} + C.$

Summary:

The solution to the given integral is: $\boxed{e^{\tan^{-1}x} + C.}$

Would you like further clarification on any step? Here are some related questions to explore:

1. What is the derivative of $\tan^{-1}x$ and how is it derived?
2. How do you integrate $e^{f(x)}$ when $f'(x)$ appears in the denominator?
3. What are the properties of inverse trigonometric functions like $\tan^{-1}x$?
4. How would the integral change if $e^{\tan^{-1}x}$ were replaced with a different function?
5. Can you generalize this method for other inverse trigonometric functions like $\sin^{-1}x$?

Tip: Always check for substitutions involving derivatives of common functions in the integrand for easier integration.