Question

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

Short answer

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

Step-by-step solution

Factor the Denominator

The first step is to factor the denominator of the given integral. The denominator is \(x^3 - x\). We can factor this as follows: \(x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)\).

Set Up Partial Fractions

We express the integrand \(\frac{1}{x(x-1)(x+1)}\) as a sum of partial fractions. This gives us: \(\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\).

Clear the Fractions

Multiply through by the common denominator \(x(x-1)(x+1)\) to clear the fractions in the equation: \[ 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1) \]

Expand and Collect Terms

Expand the right side of the equation:\[ A(x^2 - 1) + B(x^2 + x) + C(x^2 - x) \]Collect all the terms to get:\[ (A + B + C)x^2 + (B - C)x - A \]

Solve for Coefficients

Equate the coefficients of the polynomial to solve for \(A\), \(B\), and \(C\):1. \(A + B + C = 0\)2. \(B - C = 0\)3. \(-A = 1\)From equation 3, \(A = -1\). Substituting \(A = -1\) into equation 1 and 2, we get:- \(B = C\) from 2.- \(-1 + B + B = 0\) \(\Rightarrow 2B - 1 = 0\) \(\Rightarrow B = \frac{1}{2}\).- \(C = \frac{1}{2}\).

Write the Integral with Partial Fractions

Now substitute \(A\), \(B\), and \(C\) back into the original equation:\[ \frac{1}{x(x-1)(x+1)}= \frac{-1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1} \]Therefore, \(\int \frac{dx}{x(x-1)(x+1)} = \int \left(\frac{-1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1}\right) dx\).

Integrate Each Term

Integrate each term separately:\[ \int \frac{-1}{x} \, dx = -\ln|x| + C_1 \]\[ \int \frac{1/2}{x-1} \, dx = \frac{1}{2} \ln|x-1| + C_2 \]\[ \int \frac{1/2}{x+1} \, dx = \frac{1}{2} \ln|x+1| + C_3 \]

Combine the Result

Combining all the integrated terms, we get:\[ -\ln|x| + \frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| + C \]where \(C = C_1 + C_2 + C_3\) is the constant of integration.

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus that is used to find the area under a curve or the accumulation of quantities. When dealing with complex functions, direct integration might not always be straightforward. This is where various integration techniques come in handy. One such technique is known as "Partial Fraction Decomposition," which is particularly useful for integrating rational functions, where the numerator and the denominator are polynomials.

With partial fractions, we break down a complex fraction into a sum of simpler fractions. This can make integration much easier. In our example, the integral \(\int \frac{d x}{x^{3}-x}\) utilizes this technique by first factoring the denominator and rewriting the integrand as a sum of simpler fractions. Once decomposed, each term can be integrated separately using basic integration formulas, such as the integral of \(\frac{1}{x}\), which is \(\ln |x|\).

By mastering integration techniques like partial fraction decomposition, you'll be able to tackle a wide range of problems that involve complex rational functions.

Polynomial Factoring

Before applying the method of partial fractions, it's crucial to factor the polynomial completely. Factoring is about expressing a polynomial as a product of its roots or factors, which simplifies complex expressions.

For the integral \(\int \frac{d x}{x^{3}-x}\), the polynomial in the denominator is \(x^3 - x\). This can be factored step-by-step as:

• First, factor out the greatest common factor, \(x\), which gives \(x(x^2 - 1)\).
• Next, recognize that \(x^2 - 1\) is a difference of squares, \((x - 1)(x + 1)\).
• Thus, we fully factor \(x^3 - x\) as \(x(x - 1)(x + 1)\).

Factorization simplifies the polynomial and forms the basis of setting up partial fractions. This step is essential, as it translates the original complex expression into multiple simpler terms that are easier to manage.

Algebraic Fractions

Understanding and manipulating algebraic fractions is a key skill in calculus. These are expressions where one polynomial divides another, and simplifying them involves reducing the expression to its simplest form.

The principle behind partial fraction decomposition is to convert these complex algebraic fractions into a sum of simpler fractions. For the example \(\int \frac{d x}{x^{3}-x}\), the denominator \(x^3 - x\) was factored as \(x(x - 1)(x + 1)\). We then express the algebraic fraction as:

• \(\frac{A}{x}\) for the factor \(x\)
• \(\frac{B}{x-1}\) for the factor \(x-1\)
• \(\frac{C}{x+1}\) for the factor \(x+1\)

These simple fractions can be more easily integrated. By expressing any complex algebraic fraction as such sum, you simplify the integration process, turning it into a series of simple logarithmic functions. This is what makes the method of partial fractions such a powerful tool in calculus.