The given integral is:

$\int_{-\infty}^{+\infty} \frac{dx}{x^2 - 2x + 2}$

To solve this improper integral, let's first simplify the quadratic expression in the denominator.

The quadratic expression $x^2 - 2x + 2$ can be rewritten as a perfect square:

$x^2 - 2x + 2 = (x - 1)^2 + 1$

Now the integral becomes:

$\int_{-\infty}^{+\infty} \frac{dx}{(x - 1)^2 + 1}$

This is a standard form of the integral:

$\int \frac{dx}{(x - a)^2 + b^2} = \frac{1}{b} \cdot \tan^{-1}\left(\frac{x - a}{b}\right) + C$

In this case, $a = 1$ and $b = 1$, so the integral simplifies to:

$\int_{-\infty}^{+\infty} \frac{dx}{(x - 1)^2 + 1} = \tan^{-1}(x - 1) \Big|_{-\infty}^{+\infty}$

Evaluating the limits:

$\tan^{-1}(+\infty) = \frac{\pi}{2}, \quad \tan^{-1}(-\infty) = -\frac{\pi}{2}$

So the result of the integral is:

$\frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi$

Therefore, the value of the integral is:

$\boxed{\pi}$

Let me know if you'd like more details or have any questions!

Here are some related questions to explore further:

1. How does completing the square help in simplifying integrals?
2. What is the geometric interpretation of the arctangent function?
3. How are improper integrals defined and evaluated?
4. Can we use other methods like residues to solve this integral?
5. How do trigonometric substitutions work in integrals involving quadratics?

Tip: When dealing with quadratic expressions in integrals, completing the square is a powerful tool to simplify the expression.