We need to evaluate the definite integral: $\int_{-1}^{0} \frac{dx}{4x^2 + 8x + 8}$

1. Problem Description

We need to evaluate the definite integral:

$\int_{-1}^{0} \frac{dx}{4x^2 + 8x + 8}$

2. Solution Steps

First, factor out the constant 4 from the denominator:

$\int_{-1}^{0} \frac{dx}{4x^2 + 8x + 8} = \int_{-1}^{0} \frac{dx}{4(x^2 + 2x + 2)}$

Then,

$\int_{-1}^{0} \frac{dx}{4(x^2 + 2x + 2)} = \frac{1}{4} \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2}$

Complete the square for the expression in the denominator:

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

Substitute this back into the integral:

$\frac{1}{4} \int_{-1}^{0} \frac{dx}{(x+1)^2 + 1}$

Use the substitution $u = x+1$, so $du = dx$. When $x=-1$, $u = -1+1=0$. When $x=0$, $u = 0+1=1$.

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

The integral of $\frac{1}{u^2 + 1}$ is $\arctan(u)$. Thus, we have:

$\frac{1}{4} [\arctan(u)]_{0}^{1} = \frac{1}{4} (\arctan(1) - \arctan(0))$

Since $\arctan(1) = \frac{\pi}{4}$ and $\arctan(0) = 0$, we have:

$\frac{1}{4} (\frac{\pi}{4} - 0) = \frac{\pi}{16}$

3. Final Answer

$\frac{\pi}{16}$