Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1–40, use the method of partial fraction decomposition to perform the required integration. $$ \int \frac{2 x^{2}+x-4}{x^{3}-x^{2}-2 x} d x $$

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

First, we need to factor the denominator of the integrand: \( x^{3} - x^{2} - 2x \). We can factor out \( x \), giving us \( x (x^{2} - x - 2) \). Further factoring \( x^{2} - x - 2 \) yields \((x - 2)(x + 1)\). Thus, the factored form of the denominator is \( x (x - 2)(x + 1) \).

Step 2: Set Up the Partial Fraction Decomposition

Now, express the fraction \( \frac{2x^{2} + x - 4}{x (x - 2) (x + 1)} \) as a sum of partial fractions: \( \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1} \).

Step 3: Clear the Denominators

Multiply through by the common denominator \( x (x - 2)(x + 1) \) to eliminate the denominators: \[ 2x^{2} + x - 4 = A(x - 2)(x + 1) + Bx(x + 1) + Cx(x - 2) \].

Step 4: Solve for Coefficients

Expand the equation and gather like terms: - For \(A(x - 2)(x + 1)\), we have: \(Ax^2 - Ax - 2A \) - For \(Bx(x + 1)\), we have: \(Bx^2 + Bx \)- For \(Cx(x - 2)\), we have: \(Cx^2 - 2Cx \) Combine these to get: \((A + B + C)x^2 + (-A + B - 2C)x - 2A = 2x^2 + x - 4\).By comparing coefficients: - \(A + B + C = 2\) - \(-A + B - 2C = 1\) - \(-2A = -4\)Solving these equations gives: \(A = 2\), \(B = 0\), \(C = 0\).

Step 5: Integrate Each Partial Fraction

Substitute back the values of \(A\), \(B\), and \(C\):\( \int \frac{2}{x} dx \).The integral of \(\frac{2}{x}\) is: \(2 \ln |x| + C\), where \(C\) is the constant of integration.

Key Concepts

Integration Techniques

Integration is an essential tool in calculus, often used to find areas under curves or solve differential equations. One such technique is Partial Fraction Decomposition.
This method is particularly useful for integrating rational functions, such as in our original exercise.

To tackle these integrals, start by expressing the complex rational function as a sum of simpler fractions.
These simple fractions have denominators that are factors of the original denominator.

• First, factor the denominator if possible.
• The decomposition helps break down the integral into simpler parts that are easier to handle.

The integral of each simple fraction can often be found in basic integral formulae, making it easier to evaluate the overall integral.
This modular approach simplifies the process and allows for a step-by-step solution, as demonstrated in the given exercise.

Rational Functions

Rational functions are quotients of two polynomials. They play a key role in calculus and many applications of mathematics.

In the original exercise, the problem involves integrating a rational function, given by \( \int \frac{2x^{2} + x - 4}{x^{3} - x^{2} - 2x} \, dx \).
Understanding how to manipulate these functions is crucial for integration and other calculus operations.

• The numerator is a polynomial of degree 2.
• The denominator is a polynomial of degree 3.

In partial fraction decomposition, we express a complex rational function as a sum of simpler fractions.
By breaking the function down, we make it easier to understand and integrate each part.
Simple fractions typically take the form \( \frac{A}{x} \), \( \frac{B}{x-a} \), where **A** and **B** are constants to be determined.

Polynomial Factoring

Factoring is a key technique when dealing with polynomials, especially in calculus and algebra.
To decompose a rational function into partial fractions, the first step is usually to factor the polynomial in the denominator.
In our exercise, the denominator \( x^{3} - x^{2} - 2x \) is factored by:

• Factoring out the common term \( x \) to get \( x(x^{2} - x - 2) \)
• Then factoring \( x^{2} - x - 2 \) to get \( (x - 2)(x + 1) \)

This results in a completely factored denominator: \( x (x - 2)(x + 1) \).
Factoring is crucial because it identifies the simple linear factors which will be the denominators in the partial fraction decomposition.
With these factors, we set up our partial fractions: \( \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1} \), simplifying the integration process.