3.5 Q-22

Question Statement

Evaluate the integral:
$\int \frac{12}{x^3 + 8} , dx$

---

Background and Explanation

This is an integration problem involving a rational function with a cubic denominator. Recognize that the denominator can be factored as a sum of cubes: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

This suggests the use of partial fraction decomposition to break the expression into simpler components for easier integration.

---

Solution

We start by factoring the denominator: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

Thus, we can express the integral as: $\int \frac{12}{(x+2)(x^2 - 2x + 4)} , dx$

We apply partial fraction decomposition: $\frac{12}{(x+2)(x^2 - 2x + 4)} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 4} \quad \text{(i)}$

Multiplying both sides by the denominator $(x+2)(x^2 - 2x + 4)$ to clear the fractions: $12 = A(x^2 - 2x + 4) + (Bx + C)(x + 2) \quad \text{(ii)}$

Now, let’s solve for the constants $A$, $B$, and $C$:

1. Substitute $x = -2$ into equation (ii): $12 = A \left[ (-2)^2 - 2(-2) + 4 \right] + 0$ $12 = A(4 + 4 + 4)$ $12 = A(12) \quad \Rightarrow \quad A = 1$

2. Now, comparing coefficients of $x^2$ and $x$ from both sides of equation (ii):

   • From the $x^2$-terms:
     $0 = A + B \quad \Rightarrow \quad B = -1$

   • From the $x^0$-terms:
     $12 = 4A + 2C \quad \Rightarrow \quad 12 = 4(1) + 2C$ $12 = 4 + 2C \quad \Rightarrow \quad C = 4$

Thus, we have the values: $A = 1, \quad B = -1, \quad C = 4$

Now, substituting these into equation (i), we can rewrite the integral as: $\int \frac{12}{(x+2)(x^2 - 2x + 4)} , dx = \int \frac{1}{x+2} , dx - \int \frac{x - 1}{x^2 - 2x + 4} , dx$

Next, split the second term into two integrals: $= \int \frac{1}{x+2} , dx - \int \frac{x}{x^2 - 2x + 4} , dx + \int \frac{1}{x^2 - 2x + 4} , dx$

1. The first integral is straightforward: $\int \frac{1}{x+2} , dx = \ln |x+2|$

2. For the second integral, use substitution: Let $u = x^2 - 2x + 4$, so $du = (2x - 2)dx$, and rewrite the integral: $\int \frac{x}{x^2 - 2x + 4} , dx = \frac{1}{2} \ln \left( x^2 - 2x + 4 \right)$

3. The third integral involves a standard arctangent form: $\int \frac{1}{x^2 - 2x + 4} , dx = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x-1}{\sqrt{3}} \right)$

Thus, the final solution is: $\ln |x+2| - \frac{1}{2} \ln (x^2 - 2x + 4) + \sqrt{3} \tan^{-1} \left( \frac{x-1}{\sqrt{3}} \right) + C$

---

Key Formulas or Methods Used

• Sum of cubes formula:
  $x^3 + a^3 = (x + a)(x^2 - ax + a^2)$

• Partial fraction decomposition:
  Break down complex rational functions into simpler components.

• Standard integrals:

  • $\int \frac{1}{x+2} , dx = \ln |x+2|$
  • $\int \frac{1}{x^2 + a^2} , dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right)$

---

Summary of Steps

1. Factor the denominator:
   $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

2. Apply partial fraction decomposition: $\frac{12}{(x+2)(x^2 - 2x + 4)} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 4}$

3. Solve for $A$, $B$, and $C$ using substitution and comparison of coefficients: $A = 1, \quad B = -1, \quad C = 4$

4. Rewrite the integral and split into simpler integrals: $\int \frac{1}{x+2} , dx - \int \frac{x}{x^2 - 2x + 4} , dx + \int \frac{1}{x^2 - 2x + 4} , dx$

5. Integrate each term:

   • $\int \frac{1}{x+2} , dx = \ln |x+2|$
   • $\int \frac{x}{x^2 - 2x + 4} , dx = \frac{1}{2} \ln (x^2 - 2x + 4)$
   • $\int \frac{1}{x^2 - 2x + 4} , dx = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x-1}{\sqrt{3}} \right)$

6. Combine all results to get the final solution: $\ln |x+2| - \frac{1}{2} \ln (x^2 - 2x + 4) + \sqrt{3} \tan^{-1} \left( \frac{x-1}{\sqrt{3}} \right) + C$