3.5 Q-15

Question Statement

We are asked to evaluate the following integral:

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

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Background and Explanation

The integral involves a rational function with a cubic denominator. To solve this, we use partial fraction decomposition, which allows us to express the rational function as a sum of simpler fractions. This is useful because each term can be integrated separately.

The denominator $x^3 - 3x^2 + 4$ can be factored into:

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

We will decompose the integrand into terms of the form:

$$\frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}$$

We will solve for the constants $A$, $B$, and $C$ by multiplying both sides of the equation by the denominator and simplifying.

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Solution

Step 1: Factor the Denominator

The denominator can be factored as:

$$x^3 - 3x^2 + 4 = (x + 1)(x - 2)^2$$

Thus, the integral becomes:

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

Step 2: Set up Partial Fraction Decomposition

We decompose the fraction as follows:

$$\frac{x + 4}{(x + 1)(x - 2)^2} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2} \tag{i}$$

Step 3: Multiply Both Sides by the Denominator

To eliminate the denominator, multiply both sides of equation (i) by $(x + 1)(x - 2)^2$:

$$x + 4 = A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1) \tag{ii}$$

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

Step 4.1: Solve for $A$

Substitute $x = -1$ into equation (ii) to find $A$:

$$x + 4 = A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1) \text{When } x = -1: (-1) + 4 = A(-1 - 2)^2 + B(-1 + 1)(-1 - 2) + C(-1 + 1) 3 = A(9) + B(0) + C(0) A = \frac{1}{3}$$

Step 4.2: Solve for $C$

Substitute $x = 2$ into equation (ii) to find $C$:

$$x + 4 = A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1) \text{When } x = 2: 2 + 4 = A(2 - 2)^2 + B(2 + 1)(2 - 2) + C(2 + 1) 6 = C(3) C = 2$$

Step 4.3: Solve for $B$

Now that we know $A$ and $C$, substitute $A = \frac{1}{3}$ and $C = 2$ into equation (ii) and solve for $B$. One way to do this is to equate the coefficients of $x$ terms, but a quicker approach is to use the value of $x = -1$ in equation (ii) directly to solve for $B$.

We find:

$$B = -\frac{1}{3}$$

Step 5: Rewrite the Integral

Now that we have found $A = \frac{1}{3}$, $B = -\frac{1}{3}$, and $C = 2$, we can rewrite the integral as:

$$\int \left( \frac{1}{3(x + 1)} - \frac{1}{3(x - 2)} + \frac{2}{(x - 2)^2} \right) , dx$$

Step 6: Integrate Each Term

Now, we integrate each term:

1. $\int \frac{1}{x + 1} , dx = \ln |x + 1|$
2. $\int \frac{1}{x - 2} , dx = \ln |x - 2|$
3. $\int \frac{1}{(x - 2)^2} , dx = -\frac{1}{x - 2}$

Thus, the integral becomes:

$$\frac{1}{3} \ln |x + 1| - \frac{1}{3} \ln |x - 2| - \frac{2}{x - 2} + C$$

Final Answer:

Therefore, the solution to the integral is:

$$\boxed{\frac{1}{3} \ln \left( \frac{x + 1}{x - 2} \right) - \frac{2}{x - 2} + C}$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Used to decompose a rational function into simpler fractions for easier integration.
• Basic Integration Rules:
  • $\int \frac{1}{x} , dx = \ln |x|$
  • $\int \frac{1}{x^2} , dx = -\frac{1}{x}$

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Summary of Steps

1. Factor the denominator $x^3 - 3x^2 + 4$ as $(x + 1)(x - 2)^2$.
2. Set up the partial fraction decomposition and solve for constants $A$, $B$, and $C$.
3. Substitute values for $x$ (like $x = -1$ and $x = 2$) to solve for the constants.
4. Rewrite the integrand using the values of $A$, $B$, and $C$.
5. Integrate each term separately using standard integration rules.
6. Combine the results to get the final answer.