Problem 25 - Calculus for Biology and Medicine

Question

In Problems , complete the square in the denominator and evaluate the integral. $$ \int \frac{1}{x^{2}-4 x+13} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{3}\arctan\left(\frac{x-2}{3}\right) + C \).

Step 1: Identify the Denominator

The given integral is \( \int \frac{1}{x^{2}-4x+13} \, dx \). Notice that the denominator is \( x^2 - 4x + 13 \). Our goal is to complete the square for this quadratic expression.

Step 2: Completing the Square

We will complete the square for the expression \( x^2 - 4x + 13 \). Begin by taking the coefficient of \( x \), which is \( -4 \), halving it to get \( -2 \), and squaring it to get \( 4 \). Thus, we add and subtract \( 4 \) within the expression: \( x^2 - 4x + 4 + 13 - 4 = (x-2)^2 + 9 \).

Step 3: Write the Expression in Standard Form

This allows us to rewrite the denominator as \( (x-2)^2 + 9 \). Therefore, the integral becomes \( \int \frac{1}{(x-2)^2 + 9} \, dx \).

Step 4: Apply the Integration Formula

Recognize that this integral has the form \( \int \frac{1}{u^2 + a^2} \, du \), which results in \( \frac{1}{a}\arctan\left(\frac{u}{a}\right) + C \). Here, \( a^2 = 9 \) so \( a = 3 \), and \( u = x-2 \).

Step 5: Integrate

Using the formula, the integral evaluates to \( \frac{1}{3}\arctan\left(\frac{x-2}{3}\right) + C \). This is because the \( a \) in our formula equals to 3, and the substitution \( x - 2 \) (as \( u \)) matches the \( (x-2)^2 \) term.

Key Concepts

Completing the Square

Before diving into the integral itself, it's crucial to understand the process of completing the square, a technique often used in algebra and calculus. This technique transforms a quadratic expression into a perfect square trinomial, making further manipulation easier.

To complete the square, you should follow these simple steps:

• Identify the coefficient of the linear term, such as in the expression \( x^2 - 4x + 13 \), where \(-4\) is the coefficient of \(x\).
• Take half of this coefficient, \(-4/2 = -2\).
• Square the halved coefficient, \((-2)^2 = 4\), and add \(4\) to the expression to create the squared term.

By adding and subtracting \(4\), the expression becomes \((x - 2)^2 + 9\). Why does this help? Having the expression in this form allows us to apply various integration techniques easily, especially ones involving trigonometric identities.

Definite Integral

When evaluating a definite integral, you find the area under the curve of a function between two specific points. However, in many calculus problems, like the one provided, we work with indefinite integrals, which don't have specified bounds.For indefinite integrals, our focus is on finding a general solution that includes a constant of integration, often denoted as \(C\). The concept of definite integrals is crucial because:

• They have specific upper and lower bounds, providing a numerical value as the solution.
• They represent the total accumulation of the function over an interval, which is essential in areas such as physics and engineering.

While we don't solve a definite integral in this problem, understanding its fundamentals helps in appreciating the broader applicability of integrals in practical scenarios.

Trigonometric Substitution

Trigonometric substitution is a powerful technique that simplifies integrals containing certain quadratic expressions. Instead of the standard algebraic manipulation, this method involves substituting a trigonometric function that aligns naturally with the integral's structure.For example, in our problem, the denominator \((x-2)^2 + 9\) can resemble the Pythagorean identity \(u^2 + a^2\). The integrals of these forms are simplified using substitutions like \( u = a \tan \theta\).

This is why:

• It transforms the integral into a form that is easier to solve by leveraging trigonometric identities.
• It works particularly well for expressions that can be aligned with \( a^2 + b^2 \), \( a^2 - b^2 \), or similar forms.

In this exercise, bearing in mind that \( u = x - 2 \) and \( a = 3 \), the integral aligns perfectly with the formula for arctangent, neatly calculated later.

Integration Formula

Integration formulas offer blueprints for solving integrals, helping us process otherwise complex problems efficiently. For this exercise, we use the known formula for integrals of the form \(\int \frac{1}{u^2 + a^2}\, du\):

• \( \int \frac{1}{u^2 + a^2}\, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \)

This formula simplifies otherwise intricate calculations, allowing us to arrive at solutions faster. In the given problem:

• We identify \(a^2 = 9\), so \(a = 3\).
• Our substitution was \(u = x - 2\).

By plugging these into the integration formula, we achieve the result: \(\frac{1}{3}\arctan\left(\frac{x-2}{3}\right) + C\).

Using integration formulas not only saves time but also enhances understanding by connecting different mathematical concepts and techniques.