Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{x+\pi}{x^{2}-3 \pi x+2 \pi^{2}} d x\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Denominator Form

We start by examining the denominator in the integral: \(x^{2} - 3\pi x + 2\pi^{2}\). We look for factoring opportunities by treating it as a quadratic expression.

Step 2: Factoring the Denominator

Attempt to factor \(x^{2} - 3\pi x + 2\pi^{2}\) as \((x - a)(x - b)\). Solving for roots using the quadratic formula, we find that \(a = \pi\) and \(b = 2\pi\). Hence, it factors as \((x - \pi)(x - 2\pi)\).

Step 3: Setup Partial Fraction Decomposition

Express \(\frac{x + \pi}{(x - \pi)(x - 2\pi)}\) using partial fractions as \(\frac{A}{x - \pi} + \frac{B}{x - 2\pi}\).

Step 4: Solve for Coefficients

Multiply through by the denominator \((x - \pi)(x - 2\pi)\) to equate: \(x + \pi = A(x - 2\pi) + B(x - \pi)\). Expand and collect like terms to form equations: solving gives \(A = 2\) and \(B = -1\).

Step 5: Rewrite the Integral Using Partial Fractions

Substitute back into the integral: \(\int \left(\frac{2}{x - \pi} - \frac{1}{x - 2\pi}\right) dx\). Separate into two integrals: \(\int \frac{2}{x - \pi}dx - \int \frac{1}{x - 2\pi}dx\).

Step 6: Integrate Each Term

Integrate each part separately. For \(\int \frac{2}{x - \pi} dx\), the result is \(2\ln|x - \pi|\). For \(\int \frac{1}{x - 2\pi} dx\), the result is \(-\ln|x - 2\pi|\).

Step 7: Combine Results

Combine the results of the integration: \(2\ln|x - \pi| - \ln|x - 2\pi| + C\), where \(C\) is the constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus for breaking down complex rational expressions into simpler fractions that are easier to integrate.
It is particularly useful when dealing with polynomials in the denominator.

• First, you identify and factor the denominator of the rational expression, if possible.
• Next, express the original fraction as a sum of simpler fractions with unknown coefficients.
• These simpler fractions have denominators that are the factors from the original polynomial denominator.
• The numerators are constants that you solve for by equating and solving equations derived from the polynomial multiplication.

In our example, the expression \(\frac{x+\pi}{(x-\pi)(x-2\pi)}\)is decomposed into two simpler fractions,\(\frac{A}{x-\pi} + \frac{B}{x-2\pi}\).We find the values of \(A\) and \(B\) by solving an equation systeminvolving coefficient comparison.

Quadratic Factoring

Quadratic factoring involves breaking down a quadratic expression into the product of two linear expressions.
This is a critical step in setting up partial fraction decomposition as it allows the rational expression to be expressed in simpler terms.

Using the Quadratic Formula

When the quadratic cannot be easily factored by inspection, the quadratic formula is employed:\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]This determines the roots of the quadratic equation \(ax^2 + bx + c = 0\).
In our problem, we found the roots to be \(\pi\) and \(2\pi\) which allowed us to write the denominator as \((x - \pi)(x - 2\pi) \).
With this factorization, we can apply partial fractions more directly.

Definite Integral

A definite integral is the integral of a function over a specified interval, providing an area or value rather than a general formula. In contrast, our example deals with an indefinite integral, which does not operate over a range.
For definite integrals:

• You identify the limits of integration, which define the interval.
• Compute the integral using antiderivatives, and evaluate the result at the upper and lower limits.
• The result is the net area between the curve and the x-axis over the interval.

While determining definite integrals is used to find specific quantities, our task here is to determine an indefinite integral that doesn't have specific bounds.

Antiderivative

An antiderivative, or primitive function, reverses differentiation. When you find the antiderivative of a function, you're determining the original function whose derivative is the given function.

• The process of finding an antiderivative is known as integration.
• In our example, once partial fractions are decomposed, each simple fraction is integrated to find its antiderivative.
• The constants \(A\) and \(B\) help form these antiderivatives.

For instance, given \(\frac{2}{x-\pi}\), its antiderivative is \(2\ln|x-\pi|\),and \(\frac{-1}{x-2\pi}\) results in \(-\ln|x-2\pi|\).The antiderivative directly pairs with the process of differentiation, reversing the derivative to reconstruct the original function form.

Logarithmic Integration

Logarithmic integration refers to finding integrals that result in logarithmic functions. This typically arises with rational functions where the denominator is linear.
For a function of the form \(\frac{1}{x - a}\), the integral is \(\ln|x - a| + C\),where \(C\) is the integration constant.In our example, after decomposing using partial fraction decomposition, the integration results in logarithmic terms:

• \(2\ln|x - \pi|\)
• -\(\ln|x - 2\pi|\)

This approach stems from the property of integrals leading to a natural logarithm when the derivative is a constant multiple of \(\frac{1}{x-a}\). This is key in solving integrals that initially appear complex by leveraging the straightforward nature of logarithmic integration.