Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{2 x^{2}-x-20}{x^{2}+x-6} d x\)

Step-by-Step Solution

Answer

The integral is \(\ln |(x-2)(x+3)| + C\).

Step 1: Factor the Denominator

First, factor the quadratic in the denominator: \(x^2 + x - 6\). This can be factored as \((x-2)(x+3)\) based on the roots of the equation \(x^2 + x - 6 = 0\).

Step 2: Set Up Partial Fraction Decomposition

Write the fraction \(\frac{2x^2-x-20}{(x-2)(x+3)}\) as the sum of two partial fractions: \(\frac{A}{x-2} + \frac{B}{x+3}\).

Step 3: Solve for Constants A and B

Multiply through by the denominator \((x-2)(x+3)\) to eliminate the fractions: \[2x^2 - x - 20 = A(x+3) + B(x-2)\]Expand and group terms:\[2x^2 - x - 20 = (A+B)x + (3A-2B)\]By equating coefficients, solve:1. \(A + B = 2\)2. \(3A - 2B = -1\)Solving these equations gives: \(A = 1\), \(B = 1\).

Step 4: Integrate Each Term Separately

Rewrite the integral using the partial fractions:\[\int \left( \frac{1}{x-2} + \frac{1}{x+3} \right) dx\]Integrate each term:1. \(\int \frac{1}{x-2} \, dx = \ln |x-2| + C_1\)2. \(\int \frac{1}{x+3} \, dx = \ln |x+3| + C_2\)Combine into a single solution: \[\ln |x-2| + \ln |x+3| + C\]Using properties of logarithms, simplify it as:\[\ln |(x-2)(x+3)| + C\]

Step 5: Conclusion

The integral of \(\frac{2x^2-x-20}{x^2+x-6}\) is \(\ln |(x-2)(x+3)| + C\).

Key Concepts

Integration Techniques

Integration is a key concept in calculus. It is all about finding the area under a curve or solving differential problems. In this exercise, we use a special technique called partial fraction decomposition, which simplifies integrals with rational functions. A rational function is just a fraction where the numerator and denominator are polynomials.

A polynomial is a mathematical expression of one or more terms, and factoring plays a crucial role here. When you have a complicated fraction like \(\frac{2x^2-x-20}{x^2+x-6}\), direct integration is too complex. Partial fraction decomposition breaks it into simpler fractions, making it easier to integrate.

This method is practical in many calculus and engineering problems because it transforms a challenging integrand into a sum of simpler fractions. Each fraction can then be integrated individually.

Factoring Polynomials

Factoring polynomials is an essential skill in algebra and calculus. It means expressing a polynomial as a product of its simpler parts: the factors.

Here, we start by factoring the denominator \(x^2 + x - 6\). To factor a quadratic like this, we look for two numbers whose product is the constant term (-6) and whose sum is the coefficient of the linear term (1).

• The numbers +3 and -2 work because \(+3 - 2 = 1\) and \(+3 \times -2 = -6\).
• Thus, we rewrite the expression as \((x-2)(x+3)\).

Finding these factors allows us to set up partial fraction decomposition and simplifies the integration process.

Algebraic Manipulation

Algebraic manipulation allows us to solve complex mathematical problems by performing operations like addition, subtraction, multiplication, and division. In this exercise, we solve for constants \(A\) and \(B\), crucial in partial fractions.

After factoring, the expression becomes \(\frac{A}{x-2} + \frac{B}{x+3}\). We eliminate the denominators to equate the numerators:

• Multiply by \((x-2)(x+3)\): \(2x^2 - x - 20 = A(x+3) + B(x-2)\).
• Expand and rearrange: \(2x^2 - x - 20 = (A+B)x + (3A-2B)\).
• Equate coefficients: \(A + B = 2\) and \(3A - 2B = -1\).

Solving the system of equations yields \(A = 1\) and \(B = 1\). This manipulation is critical to progress towards integration.

Logarithm Properties

Logarithms are the inverse of exponentials. They play a vital role in integration, especially when dealing with partial fractions. After solving for constants \(A\) and \(B\), integrate each fraction separately.

When you integrate \(\int \frac{1}{x-a} \, dx\), the result is \(\ln |x-a| + C\). Thus, \(\int \left( \frac{1}{x-2} + \frac{1}{x+3} \right) dx\) becomes two separate logarithmic expressions:

• \(\int \frac{1}{x-2} \, dx = \ln |x-2| + C_1\)
• \(\int \frac{1}{x+3} \, dx = \ln |x+3| + C_2\)

Combine these using the property \(\ln a + \ln b = \ln (ab)\), simplifying the solution to \(\ln |(x-2)(x+3)| + C\).

This property not only simplifies expressions, but it is a common tool in solving differential equations and calculus problems efficiently.