Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of completing the square, along with a trigonometric substitution if needed, to evaluate each integral. \(\int \frac{2 x-1}{x^{2}-6 x+18} d x\)

Step-by-Step Solution

Answer

\(\ln |(x-3)^2 + 9| + \frac{5}{3} \tan^{-1}(\frac{x-3}{3}) + C\)

Step 1: Identify the Quadratic in the Denominator

The given integral is \( \int \frac{2x-1}{x^2 - 6x + 18} \, dx \). Notice that the denominator is a quadratic expression \(x^2 - 6x + 18\). The first step is to complete the square for the quadratic in the denominator.

Step 2: Complete the Square

To complete the square for \(x^2 - 6x + 18\), start by focusing on the terms \(x^2 - 6x\). Take half of the coefficient of \(x\), which is \(-6\), that gives \(-3\). Square it to obtain \(9\). Rewrite the quadratic as \((x-3)^2 + 18 - 9 = (x-3)^2 + 9\). So, the expression becomes \((x-3)^2 + 9\).

Step 3: Rewrite the Integral with Substitution

Now, rewrite the integral \(\int \frac{2x-1}{(x-3)^2 + 9} \, dx\). Use the substitution \(u = x - 3\), which implies \( du = dx \) and \(x = u + 3\). Substitute into the integral to get \(\int \frac{2(u+3)-1}{u^2+9} \, du\). This simplifies to \(\int \frac{2u + 5}{u^2 + 9} \, du\).

Step 4: Split the Integral

Split the integral \(\int \frac{2u + 5}{u^2 + 9} \, du\) into two separate integrals: \(\int \frac{2u}{u^2 + 9} \, du + \int \frac{5}{u^2 + 9} \, du\). The first integral can be solved using a simple substitution, and the second one involves a trigonometric substitution.

Step 5: Solve the First Integral

Let \(v = u^2 + 9\), then \(dv = 2u \, du\). The first integral becomes \( \int \frac{2u}{u^2 + 9} \, du = \int \frac{1}{v} \, dv\), which simplifies to \(\ln|v| + C_1 = \ln|u^2 + 9| + C_1\).

Step 6: Solve the Second Integral with Trigonometric Substitution

For \(\int \frac{5}{u^2 + 9} \, du\), use the substitution \(u = 3\tan(\theta)\), which implies \(du = 3\sec^2(\theta) \, d\theta\). The integral becomes \(\int \frac{5}{9\sec^2(\theta)} \times 3\sec^2(\theta) \, d\theta = \int \frac{15}{9} \, d\theta = \int \frac{5}{3} \, d\theta\). This results in \(\frac{5}{3}\theta + C_2\). Back substitute to get \(\theta = \tan^{-1}(\frac{u}{3})\), so this part of the integral is \(\frac{5}{3}\tan^{-1}(\frac{u}{3})\).

Step 7: Combine the Solutions

Combine the solutions from both integrals. We have \(\ln|u^2 + 9| + \frac{5}{3} \tan^{-1}(\frac{u}{3}) + C\). Recall that \(u = x - 3\), so the expression becomes \( \ln |(x-3)^2 + 9| + \frac{5}{3} \tan^{-1}(\frac{x-3}{3}) + C \).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This technique is particularly useful for integration, especially when dealing with irrational or trigonometric expressions.
For the quadratic expression in the denominator of the integral, \(x^2 - 6x + 18\), completing the square helps simplify the expression.

• Identify the quadratic term \(x^2 - 6x\).
• Find half of the coefficient of \(x\), which in this case is \(-3\), and then square it to get \(9\).
• Rewrite \(x^2 - 6x + 18\) as \((x-3)^2 + 18 - 9 = (x-3)^2 + 9\).

As you can see, the quadratic is now rewritten as a perfect square, which makes subsequent steps in solving the integral more straightforward.
Completing the square thus simplifies the process of integration, especially when that expression is under the sign of an integral.

Trigonometric Substitution

Trigonometric substitution is a technique in calculus used to evaluate integrals containing quadratic expressions under the square root or in the denominator.
This method involves substituting a trigonometric function for a variable to simplify the integral.
For the second integral \(\int \frac{5}{u^2 + 9} \, du\), a trigonometric substitution makes it easier.

• Set \(u = 3\tan(\theta)\), implying \(du = 3\sec^2(\theta) \, d\theta\).
• The denominator becomes \(9\sec^2(\theta)\), simplifying the integral expression.
• The integral turns into a much simpler form, \( \int \frac{5}{3} \, d\theta \).

This results in an integral with respect to \(\theta\), which is elementary to solve, leading to \(\frac{5}{3} \theta + C\).
Understanding when and how to use trigonometric substitution is critical to solving integrals that include quadratic expressions.

Definite Integrals

A definite integral has specified limits of integration and computes the net area between the curve \(f(x)\) and the x-axis over an interval \([a, b]\).
While the example provided involves an indefinite integral, recognizing when to transition to definite integration is important.
Definite integrals are particularly useful in scenarios where the exact area calculation is needed instead of a general formula.
Integrals such as \( \int_a^b f(x) \, dx \) require evaluating the antiderivative \(F(x)\) at the specific limits, \(F(b) - F(a)\).

• This sort of integration is fundamental for physical and statistical applications where precise area, total distance, or averages are required.
• Unlike indefinite integrals, definite integrals yield a numeric value, representing physical quantities such as mass or volume.

Once you grasp indefinite integration, transitioning to definite integrals involves understanding the application of limits.

Partial Fractions

Partial fractions provide a method for breaking down complex rational expressions into simpler components that are easier to integrate.
This method is particularly useful for integrals involving polynomials in the numerator and denominator.
Although the provided solution doesn't directly apply partial fractions, understanding it enhances comprehension of integral solutions.

• Break a rational function \(\frac{P(x)}{Q(x)}\) into simpler fractions like \(\frac{A}{x-r}\) or \(\frac{Bx+C}{x^2+px+q}\).
• Determine constants \(A, B, C, ...\) by setting up systems of equations based on coefficients.
• This method simplifies the integration of each fraction separately, often leading to basic natural log or arctangent functions as results.

Recognizing when to decompose an integrand using partial fractions is crucial when faced with non-linear polynomial divisions.
This technique greatly simplifies the integration process, especially when dealing with higher-order polynomials.