We need to evaluate the definite integral: $\int_{-1}^{\infty} \frac{dx}{4x^2+8x+8}$.

1. Problem Description

We need to evaluate the definite integral:

$\int_{-1}^{\infty} \frac{dx}{4x^2+8x+8}$.

2. Solution Steps

First, we can factor out the constant 4 from the denominator:

$\int_{-1}^{\infty} \frac{dx}{4(x^2+2x+2)}$

$= \frac{1}{4} \int_{-1}^{\infty} \frac{dx}{x^2+2x+2}$.

Next, we complete the square in the denominator:

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

So the integral becomes:

$\frac{1}{4} \int_{-1}^{\infty} \frac{dx}{(x+1)^2+1}$.

Let $u = x+1$, then $du = dx$. When $x = -1$, $u = -1+1 = 0$. When $x \to \infty$, $u \to \infty$.

So the integral becomes:

$\frac{1}{4} \int_{0}^{\infty} \frac{du}{u^2+1}$.

We know that $\int \frac{du}{u^2+1} = \arctan(u) + C$, so:

$\frac{1}{4} \int_{0}^{\infty} \frac{du}{u^2+1} = \frac{1}{4} [\arctan(u)]_{0}^{\infty}$.

As $u$ approaches infinity, $\arctan(u)$ approaches $\frac{\pi}{2}$. Also, $\arctan(0) = 0$.

Therefore,

$\frac{1}{4} [\arctan(u)]_{0}^{\infty} = \frac{1}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8}$.

3. Final Answer

The final answer is $\frac{\pi}{8}$.