To integrate the function $\frac{x^2 + 8x + 128}{x^3 - 64x}$, we can first attempt to simplify the expression, if possible, and then apply integration techniques.

Step 1: Factor the denominator

The denominator $x^3 - 64x$ can be factored as follows: $x^3 - 64x = x(x^2 - 64) = x(x - 8)(x + 8)$

So the integral becomes: $\int \frac{x^2 + 8x + 128}{x(x - 8)(x + 8)} \, dx$

Step 2: Use partial fraction decomposition

We can express the fraction as a sum of simpler fractions: $\frac{x^2 + 8x + 128}{x(x - 8)(x + 8)} = \frac{A}{x} + \frac{B}{x - 8} + \frac{C}{x + 8}$ We will now solve for $A$, $B$, and $C$.

Step 3: Set up the equation for partial fractions

Multiply both sides by the denominator $x(x - 8)(x + 8)$ to eliminate the denominators: $x^2 + 8x + 128 = A(x - 8)(x + 8) + Bx(x + 8) + Cx(x - 8)$

Expand both sides: $x^2 + 8x + 128 = A(x^2 - 64) + Bx(x + 8) + Cx(x - 8)$ Simplifying both sides: $x^2 + 8x + 128 = A(x^2 - 64) + B(x^2 + 8x) + C(x^2 - 8x)$ $x^2 + 8x + 128 = (A + B + C)x^2 + (8B - 8C)x - 64A$

Step 4: Solve the system of equations

By comparing the coefficients of $x^2$, $x$, and the constant term on both sides, we get the system: $A + B + C = 1 \quad \text{(coefficient of } x^2 \text{)}$ $8B - 8C = 8 \quad \text{(coefficient of } x \text{)}$ $-64A = 128 \quad \text{(constant term)}$

Solve the system:

1. From the constant term: $-64A = 128$ ⟹ $A = -2$.
2. Substitute $A = -2$ into $A + B + C = 1$: $-2 + B + C = 1 \quad \Rightarrow \quad B + C = 3$
3. From $8B - 8C = 8$, divide both sides by 8: $B - C = 1$

Now solve the system: $B + C = 3 \quad \text{and} \quad B - C = 1$ Add the two equations: $2B = 4 \quad \Rightarrow \quad B = 2$ Substitute $B = 2$ into $B + C = 3$: $2 + C = 3 \quad \Rightarrow \quad C = 1$

Step 5: Write the partial fraction decomposition

Now that we have $A = -2$, $B = 2$, and $C = 1$, the original expression becomes: $\frac{x^2 + 8x + 128}{x(x - 8)(x + 8)} = \frac{-2}{x} + \frac{2}{x - 8} + \frac{1}{x + 8}$

Step 6: Integrate term by term

Now integrate each term separately: $\int \frac{-2}{x} \, dx = -2 \ln |x|$ $\int \frac{2}{x - 8} \, dx = 2 \ln |x - 8|$ $\int \frac{1}{x + 8} \, dx = \ln |x + 8|$

Thus, the integral is: $\int \frac{x^2 + 8x + 128}{x(x - 8)(x + 8)} \, dx = -2 \ln |x| + 2 \ln |x - 8| + \ln |x + 8| + C$

Final Answer:

$\boxed{-2 \ln |x| + 2 \ln |x - 8| + \ln |x + 8| + C}$

Do you have any questions, or would you like further details on a particular step?

Here are 5 related questions:

1. How do you apply partial fraction decomposition in more complex rational expressions?
2. What is the role of logarithmic functions in integrals involving fractions?
3. How do you identify when to use partial fractions versus other methods of integration?
4. Can you always factor cubic expressions like $x^3 - 64x$?
5. How would you approach integrating higher-degree polynomials over factored terms?

Tip: Whenever you encounter a rational function, check if partial fraction decomposition can simplify the integration process!