3.5 Q-23

Question Statement

Evaluate the integral:

$$\int \frac{9x + 6}{x^2 - 8} , dx$$

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Background and Explanation

This is an integral involving a rational function with a quadratic denominator. To solve it, we will decompose the expression using partial fraction decomposition. The denominator $x^2 - 8$ can be factored into $(x - 2)(x^2 + 2x + 4)$. The method we will use is to express the integrand as a sum of simpler fractions that can be integrated individually.

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Solution

Step 1: Factor the denominator

We start by recognizing the factored form of the denominator:

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

Thus, the integral becomes:

$$\int \frac{9x + 6}{(x - 2)(x^2 + 2x + 4)} , dx$$

Step 2: Apply partial fraction decomposition

We assume that the integrand can be decomposed into the form:

$$\frac{9x + 6}{(x - 2)(x^2 + 2x + 4)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + 2x + 4}$$

Step 3: Multiply through by the common denominator

To eliminate the denominators, multiply both sides of the equation by $(x - 2)(x^2 + 2x + 4)$:

$$9x + 6 = A(x^2 + 2x + 4) + (Bx + C)(x - 2)$$

Step 4: Expand both sides

Now, expand both sides of the equation:

$$9x + 6 = A(x^2 + 2x + 4) + (Bx + C)(x - 2)$$

Expanding the right-hand side:

$$9x + 6 = A(x^2 + 2x + 4) + Bx(x - 2) + C(x - 2)$$

This simplifies to:

$$9x + 6 = A(x^2 + 2x + 4) + Bx^2 - 2Bx + Cx - 2C$$

Step 5: Set up the system of equations

Now, collect terms:

$$9x + 6 = (A + B)x^2 + (2A - 2B + C)x + (4A - 2C)$$

Equate the coefficients of like powers of $x$ on both sides:

1. For $x^2$: $A + B = 0$
2. For $x$: $2A - 2B + C = 9$
3. For the constant term: $4A - 2C = 6$

Step 6: Solve the system of equations

From equation (1):

$$A + B = 0 \quad \Rightarrow \quad B = -A$$

Substitute $B = -A$ into the other two equations:

From equation (2):

$$2A - 2(-A) + C = 9 \quad \Rightarrow \quad 4A + C = 9$$

From equation (3):

$$4A - 2C = 6$$

Now, solve these two equations:

1. $4A + C = 9$
2. $4A - 2C = 6$

Multiply the first equation by 2:

$$8A + 2C = 18$$

Now, add it to the second equation:

$$(8A + 2C) + (4A - 2C) = 18 + 6$$

This simplifies to:

$$12A = 24 \quad \Rightarrow \quad A = 2$$

Now, substitute $A = 2$ into $B = -A$:

$$B = -2$$

Finally, substitute $A = 2$ into $4A + C = 9$:

$$4(2) + C = 9 \quad \Rightarrow \quad 8 + C = 9 \quad \Rightarrow \quad C = 1$$

Step 7: Rewrite the integral

Now that we have the values for $A$, $B$, and $C$, we can rewrite the partial fraction decomposition:

$$\frac{9x + 6}{(x - 2)(x^2 + 2x + 4)} = \frac{2}{x - 2} + \frac{-2x + 1}{x^2 + 2x + 4}$$

Thus, the integral becomes:

$$\int \frac{9x + 6}{(x - 2)(x^2 + 2x + 4)} , dx = 2 \int \frac{dx}{x - 2} - \int \frac{2x + 2 - 3}{x^2 + 2x + 4} , dx$$

Step 8: Integrate each term

The first term is straightforward:

$$2 \int \frac{dx}{x - 2} = 2 \ln |x - 2|$$

For the second term, split it into two parts:

$$-\int \frac{2x + 2}{x^2 + 2x + 4} , dx + 3 \int \frac{1}{x^2 + 2x + 4} , dx$$

For the first part:

$$-\int \frac{2x + 2}{x^2 + 2x + 4} , dx = -\ln(x^2 + 2x + 4)$$

For the second part, use the standard integral for arctangent:

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

Step 9: Combine the results

Finally, combine all the results:

$$\int \frac{9x + 6}{(x - 2)(x^2 + 2x + 4)} , dx = 2 \ln |x - 2| - \ln(x^2 + 2x + 4) + \sqrt{3} \tan^{-1}\left( \frac{x - 1}{\sqrt{3}} \right) + C$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Decompose the rational function into simpler fractions to simplify integration.
• Basic Integration Formulas:
  • $\int \frac{dx}{x - a} = \ln |x - a| + C$
  • $\int \frac{dx}{x^2 + ax + b} = \frac{1}{\sqrt{b}} \tan^{-1} \left( \frac{x - h}{\sqrt{b}} \right) + C$

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Summary of Steps

1. Factor the denominator as $(x - 2)(x^2 + 2x + 4)$.
2. Decompose the integrand using partial fractions.
3. Solve for constants $A$, $B$, and $C$ using a system of equations.
4. Rewrite the integral in simpler parts.
5. Integrate each part separately.
6. Combine the results to get the final answer.