1. The problem is to evaluate the improper integral $$\int_{-\infty}^{\infty} \frac{1}{x^2 + 2x + 2} \, dx.$$\n\n2. First, complete the square in the denominator: $$x^2 + 2x + 2 = (x+1)^2 + 1.$$\n\n3. The integral becomes $$\int_{-\infty}^{\infty} \frac{1}{(x+1)^2 + 1} \, dx.$$\n\n4. Use the substitution $u = x + 1$, so $du = dx$. The limits remain $-\infty$ to $\infty$. The integral is now $$\int_{-\infty}^{\infty} \frac{1}{u^2 + 1} \, du.$$\n\n5. Recall the standard integral formula: $$\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C.$$ Here, $a = 1$.\n\n6. Evaluate the definite integral: $$\int_{-\infty}^{\infty} \frac{1}{u^2 + 1} \, du = \left[ \arctan(u) \right]_{-\infty}^{\infty} = \arctan(\infty) - \arctan(-\infty) = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi.$$\n\n7. Therefore, the value of the integral is $$\boxed{\pi}.$$