Problem 32 - Calculus Single Variable

Question

Use the method of partial fractions to calculate the given integral. $$ \int \frac{7 x^{2}+7 x-2}{x(x-1)(x+2)} d x $$

Step-by-Step Solution

Answer

The integral is \( \ln |x| + 4 \ln |x-1| + 2 \ln |x+2| + C \).

Step 1: Understand the problem

We need to evaluate the integral \( \int \frac{7x^{2} + 7x - 2}{x(x-1)(x+2)} \, dx \) using the method of partial fractions. This method involves expressing the integrand as a sum of simpler fractions that can be integrated easily.

Step 2: Decompose the fraction

Express \( \frac{7x^{2} + 7x - 2}{x(x-1)(x+2)} \) as a sum of partial fractions. Assume: \[ \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+2} = \frac{7x^{2} + 7x - 2}{x(x-1)(x+2)} \] Multiply through by the denominator \(x(x-1)(x+2)\) to clear the fractions.

Step 3: Clear the denominators

After multiplying through, we have: \[ A(x-1)(x+2) + Bx(x+2) + Cx(x-1) = 7x^{2} + 7x - 2 \]Expand each term on the left side to align with the right side.

Step 4: Equate and solve for coefficients

Expand and collect like terms: \[ A(x^2 + x - 2) + B(x^2 + 2x) + C(x^2 - x) = 7x^{2} + 7x - 2 \] This results in the equation: \[ (A + B + C)x^2 + (A + 2B - C)x - 2A = 7x^2 + 7x - 2 \] Match coefficients to form the system: 1. \( A + B + C = 7 \)2. \( A + 2B - C = 7 \)3. \( -2A = -2 \)

Step 5: Solve the system of equations

Solve the equations: From equation 3: \( -2A = -2 \Rightarrow A = 1 \) Substitute \( A = 1 \) into equations 1 and 2:1. \( 1 + B + C = 7 \Rightarrow B + C = 6 \)2. \( 1 + 2B - C = 7 \Rightarrow 2B - C = 6 \)Solve this system: Find \( B \) by solving \( B + C = 6 \) and \( 2B - C = 6 \):Add these equations: \( 3B = 12 \Rightarrow B = 4 \)Now substitute \( B = 4 \) into \( B + C = 6 \) to find \( C = 2 \).

Step 6: Write the partial fraction expression

With the coefficients, the partial fraction decomposition becomes: \[ \frac{1}{x} + \frac{4}{x-1} + \frac{2}{x+2} \] Thus, the integral becomes: \[ \int \left( \frac{1}{x} + \frac{4}{x-1} + \frac{2}{x+2} \right) \, dx \]

Step 7: Integrate the partial fractions

Integrate each term separately: \[ \int \frac{1}{x} \, dx = \ln |x| + C_1 \] \[ \int \frac{4}{x-1} \, dx = 4 \ln |x-1| + C_2 \] \[ \int \frac{2}{x+2} \, dx = 2 \ln |x+2| + C_3 \] Combine, including the constant of integration, \( C \): \[ \ln |x| + 4 \ln |x-1| + 2 \ln |x+2| + C \]

Step 8: Final Answer

The integral is: \[ \ln |x| + 4 \ln |x-1| + 2 \ln |x+2| + C \]

Key Concepts

Understanding Integrals

Integrals are a fundamental concept in calculus. They can be thought of as the reverse process of differentiation. While differentiation deals with the rate of change, integration focuses on accumulation – this means finding the total amount or area under a curve. The integral in this exercise \[ \int \frac{7 x^{2}+7 x-2}{x(x-1)(x+2)} \, dx \]represents the area under the curve of the given function.

To solve integrals of complex rational functions, one useful technique is the method of partial fractions. By breaking down a complicated fraction into simpler parts, we can find the antiderivative more easily. This step-by-step decomposition helps us tackle the integral by treating each simpler fraction separately, making our calculations more manageable.

Breaking Down Rational Functions

Rational functions are the quotient of two polynomials. In our example, \[ \frac{7 x^{2}+7 x-2}{x(x-1)(x+2)} \]we have a rational function where the numerator is a polynomial of degree 2, and the denominator is a product of linear factors.

• The first step in dealing with such integrals is to decompose the complicated rational function into simpler partial fractions.
• This involves expressing the initial function as a sum of simpler fractions, each with a different linear factor in the denominator.
• By doing this, each simpler fraction can be handled individually, making the integration process much simpler.

This decomposition is key to handling otherwise daunting integrals and allows us to apply straightforward calculus rules to solve each part.

The Role of Calculus in Partial Fractions

Calculus provides the essential tools for solving integrals. Through differentiation and integration, we can understand the behavior and properties of functions. When using partial fractions, calculus helps in several ways:

• After decomposing a rational function into partial fractions, calculus rules like the basic logarithmic integration can be applied.
• For instance, the integral of \( \frac{1}{x} \) results in \( \ln |x| \). Similarly, each partial fraction can be integrated using the natural logarithm rule.
• Calculus also allows us to verify our results by differentiating the obtained expression to ensure it matches the original function, confirming the accuracy of our integration process.

The beauty of calculus lies in its methodical approach to breaking down complex problems like partial fraction integrals into simpler, manageable tasks.