We are given the following integral:

$\int \frac{x + 8}{2x^3 - 8x} \, dx$

Step 1: Factor the denominator

First, we factor the denominator $2x^3 - 8x$:

$2x^3 - 8x = 2x(x^2 - 4) = 2x(x - 2)(x + 2)$

Thus, the integrand becomes:

$\frac{x + 8}{2x(x - 2)(x + 2)}$

Step 2: Set up partial fractions

We express the rational function as a sum of partial fractions of the form:

$\frac{x + 8}{2x(x - 2)(x + 2)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$

Step 3: Clear the denominator

Multiply both sides of the equation by $2x(x - 2)(x + 2)$ to eliminate the denominators:

$x + 8 = A \cdot 2(x - 2)(x + 2) + B \cdot 2x(x + 2) + C \cdot 2x(x - 2)$

Step 4: Solve for constants $A$, $B$, and $C$

Now, expand and collect like terms to find the values of $A$, $B$, and $C$.

For $x = 0$:

$x + 8 = A \cdot 2(-2)(2) + B \cdot 2(0)(2) + C \cdot 2(0)(-2)$ $8 = -8A$ $A = -1$

For $x = 2$:

$x + 8 = A \cdot 2(0)(4) + B \cdot 2(2)(4) + C \cdot 2(2)(0)$ $10 = 16B$ $B = \frac{10}{16} = \frac{5}{8}$

For $x = -2$:

$x + 8 = A \cdot 2(-4)(0) + B \cdot 2(-2)(0) + C \cdot 2(-2)(-4)$ $6 = 16C$ $C = \frac{6}{16} = \frac{3}{8}$

Step 5: Write the partial fraction decomposition

Now that we have the values of $A$, $B$, and $C$, we can express the integrand as:

$\frac{x + 8}{2x(x - 2)(x + 2)} = \frac{-1}{x} + \frac{5/8}{x - 2} + \frac{3/8}{x + 2}$

Final Answer

The integral is now in the form of a sum of partial fractions:

$\frac{-1}{x} + \frac{5/8}{x - 2} + \frac{3/8}{x + 2}$

Would you like to proceed with evaluating the integrals, or do you need further explanation on this part?