Question

Use the method of partial fractions to evaluate each of the following integrals. \(\int \frac{d x}{x^{3}-2 x^{2}-4 x+8}\)

Short answer

The integral is \( -\frac{1}{9} \ln|x-1| + \frac{1}{18} \ln|x-4| + \frac{1}{18} \ln|x+2| + C \).

Step-by-step solution

Factor the Denominator

The first step in using partial fractions is to factor the denominator. The polynomial in the denominator is \( x^3 - 2x^2 - 4x + 8 \). We look for a factor by using the Rational Root Theorem, which suggests trying \( x = 1 \) because it's a root. Dividing the polynomial by \( x - 1 \) gives \( (x - 1)(x^2 - x - 8) \). The quadratic \( x^2 - x - 8 \) can be factored further into \( (x - 4)(x + 2) \). Thus, the complete factorization is \( (x - 1)(x - 4)(x + 2) \).

Set Up Partial Fractions

Write the integrand as a sum of partial fractions: \( \frac{1}{(x-1)(x-4)(x+2)} = \frac{A}{x-1} + \frac{B}{x-4} + \frac{C}{x+2} \).

Clear the Denominator

Multiply both sides by the common denominator \((x - 1)(x - 4)(x + 2)\) to clear the fraction. This results in \(1 = A(x-4)(x+2) + B(x-1)(x+2) + C(x-1)(x-4)\).

Solve for Coefficients

To find the values of \( A \), \( B \), and \( C \), substitute convenient values for \( x \):- Let \( x = 1 \), then \( 1 = A(1-4)(1+2) \Rightarrow 1 = -9A \Rightarrow A = -\frac{1}{9} \)- Let \( x = 4 \), then \( 1 = B(4-1)(4+2) \Rightarrow 1 = 18B \Rightarrow B = \frac{1}{18} \)- Let \( x = -2 \), then \( 1 = C(-2-1)(-2-4) \Rightarrow 1 = 18C \Rightarrow C = \frac{1}{18} \).

Substitute Coefficients Back

Substitute the values \( A = -\frac{1}{9} \), \( B = \frac{1}{18} \), \( C = \frac{1}{18} \) into the partial fraction decomposition: \( \frac{-1}{9(x-1)} + \frac{1}{18(x-4)} + \frac{1}{18(x+2)} \).

Integrate Each Term

Integrate each term separately:- \( \int \frac{-1}{9(x-1)} \, dx = -\frac{1}{9} \ln|x-1| + C_1 \)- \( \int \frac{1}{18(x-4)} \, dx = \frac{1}{18} \ln|x-4| + C_2 \)- \( \int \frac{1}{18(x+2)} \, dx = \frac{1}{18} \ln|x+2| + C_3 \).Combine these using a single constant \( C \).

Combine Results

Combine the integrals: \[ \int \frac{dx}{x^3 - 2x^2 - 4x + 8} = -\frac{1}{9} \ln|x-1| + \frac{1}{18} \ln|x-4| + \frac{1}{18} \ln|x+2| + C \].

Key concepts

Polynomial Factorization

In partial fraction integration, polynomial factorization is crucial for breaking down complex expressions. Here, we start with the polynomial \(x^3 - 2x^2 - 4x + 8\), which we need to factor to use partial fractions effectively. The goal of polynomial factorization is to express a polynomial as a product of its simpler parts, or factors, which can be linear (first degree) or quadratic (second degree).

• First, we employ a technique known as synthetic division or traditional polynomial division to simplify the polynomial by discovering its roots.
• The Rational Root Theorem guides us in finding roots, suggesting possible rational solutions based on the factors of the constant and leading coefficient.
• In this case, we find \(x = 1\) is a root of the polynomial, allowing us to divide \(x^3 - 2x^2 - 4x + 8\) by \(x - 1\), resulting in \((x - 1)(x^2 - x - 8)\).
• The quadratic \(x^2 - x - 8\) is further factored into \((x - 4)(x + 2)\) using the method of looking for two numbers that multiply to \(-8\) and add to \(-1\).

By understanding these steps, we can seamlessly transition to setting up partial fractions for integration.

Rational Root Theorem

The Rational Root Theorem is a helpful mathematical tool in finding the potential rational roots of a polynomial equation, which ultimately aids factorization. This theorem states that any rational solution, or root, of the polynomial equation must be a fraction \(\frac{p}{q}\), where \(p\) is a factor of the constant term, and \(q\) is a factor of the leading term coefficient.

• In the polynomial \(x^3 - 2x^2 - 4x + 8\), the constant term is \(8\), and the leading coefficient is \(1\).
• The factors of 8 are \(\pm1, \pm2, \pm4, \pm8\), while the factors of 1 are \(\pm1\).
• Thus, the possible rational roots are \(\pm1, \pm2, \pm4, \pm8\).

By testing these values, we find that \(x = 1\) is indeed a root, helping us begin the factorization process necessary for setting up partial fractions.

Logarithmic Integration

When integrating functions with linear denominators, the antiderivatives often involve logarithmic functions. Each term in a partial fraction decomposition, such as \(\frac{A}{x-a}\), is integrated using the natural logarithm. Here's how it works:

• Given a fraction \(\frac{A}{x-a}\), the integral becomes \(A \ln|x-a| + C\), where \(C\) is the constant of integration.
• This approach applies when the degree of the denominator is greater than the degree of the numerator and when the denominator is linear.

For the current exercise, integrating each term of the partial fractions results in expressions like \(-\frac{1}{9} \ln|x-1|\), \(\frac{1}{18} \ln|x-4|\), and \(\frac{1}{18} \ln|x+2|\), all of which exploit this property of logarithmic integration.

Definite and Indefinite Integrals

In calculus, integrals are classified into definite and indefinite types. Understanding both is important for solving integration problems with partial fractions and other methods.

• An indefinite integral is a function that represents the family of all antiderivatives of the original function. It includes a constant \(C\) representing any arbitrary constant value. This is expressed typically as \(\int f(x) \, dx = F(x) + C\).
• In contrast, a definite integral calculates the area under the curve of a function between two specified limits \(a\) and \(b\). It's expressed as \(\int_{a}^{b} f(x) \, dx\) and results in a numerical value.

The problem at hand demonstrates indefinite integration, using partial fractions to decompose and solve the integral without specified limits, leading to a general solution with a constant \(C\).