Question

Use the method of partial fractions to evaluate the following integrals. \(\int \frac{3 x+4}{x^{3}-2 x-4} d x\) (Hint: Use the rational root theorem.)

Short answer

The integral is \(\ln|x-2| + \ln|x^2+2x+2| - \tan^{-1}(x+1) + C\).

Step-by-step solution

Factor the Denominator

Using the rational root theorem, test possible rational roots of the denominator polynomial \(x^3 - 2x - 4\). The possible rational roots are \( \pm 1, \pm 2, \pm 4\). By testing these, we find that \( x = 2 \) is a root. Thus, we can factor the polynomial as \( (x-2)(x^2 + 2x + 2) \).

Decompose into Partial Fractions

Express the integrand \( \frac{3x+4}{(x-2)(x^2+2x+2)} \) as partial fractions: \( \frac{A}{x-2} + \frac{Bx + C}{x^2 + 2x + 2} \).

Find the Coefficients

Multiply both sides by the common denominator \((x-2)(x^2+2x+2)\), then equate the numerators: \[ 3x + 4 = A(x^2 + 2x + 2) + (Bx + C)(x-2). \]Expand and collect like terms to solve the system of equations for \(A\), \(B\), and \(C\). We find that \(A = 1\), \(B = 2\), and \(C = -1\).

Integrate Each Term Separately

Now integrate the separate fractions:1. \( \int \frac{1}{x-2} \, dx = \ln|x-2| + C_1 \)2. \( \int \frac{2x-1}{x^2+2x+2} \, dx \). For the second integral, perform a substitution. Let \(u = x^2 + 2x + 2\), so \(du = (2x+2)\ dx\). Rewrite and integrate to find the result.

Combine the Integrals

Combine the results of the separate integrals to write the final solution. The integral evaluates to \[ \ln|x-2| + \ln|u| - \int \frac{1}{u} dx = \ln|x-2| + \ln(x^2+2x+2) - \tan^{-1}(x+1) + C. \]

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions. This technique is particularly helpful when integrating rational functions, which are ratios of polynomials. The idea is to break down a complex fraction into several easier parts that can be individually integrated.

To begin, the rational function's denominator is factored if possible. Factoring transforms the denominator into a product of linear and irreducible quadratic terms. For example, in the integrand \( \frac{3x+4}{x^3-2x-4}\), the denominator \(x^3-2x-4\) is factored into \((x-2)(x^2+2x+2)\).

Once factored, we express the integrand as a combination of fractions. Each fraction corresponds to a factor in the denominator. In this case, we write the expression as \( \frac{A}{x-2} + \frac{Bx+C}{x^2+2x+2} \). The coefficients (A, B, and C) are determined by equating and expanding the numerators, then solving a system of linear equations. This step is crucial as it sets up the terms for integration.

Rational Root Theorem

The Rational Root Theorem is a tool for finding the possible rational roots of a polynomial. This theorem states that any rational root of a polynomial equation \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0\) must be of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\), and \(q\) is a factor of the leading coefficient \(a_n\).

This theorem simplifies the process of locating roots by providing a finite list of candidates to test. For the exercise polynomial \(x^3 - 2x - 4\), we consider \( \pm 1\), \( \pm 2\), and \( \pm 4\) because the constant term is \(-4\) and the leading coefficient is \(1\). After testing, \(x=2\) is confirmed as a root.

Once a root is found, polynomial division can be used to further factor the polynomial. This makes solving the integral possible through simpler partial fractions.

Integral Evaluation

Evaluating integrals involves finding the antiderivative of a function. In our context, after applying partial fraction decomposition, we separately integrate each simpler fraction.

The first term, \( \frac{1}{x-2} \), is straightforward. Its integral is \( \ln|x-2| \). The second term, \( \frac{2x-1}{x^2+2x+2} \), requires a substitution. By letting \( u = x^2 + 2x + 2 \), derivative \( du = (2x+2) dx \) allows us to convert it into a family of terms easier to integrate.

This term transforms into a combination of simpler integrals that may involve log functions or inverse trigonometric functions, such as \( \tan^{-1} \). Once each part is integrated, the solutions are combined to obtain the result for the original function. This step is key to handling complex rational functions.