Problem 22 - Calculus for Biology and Medicine

Question

In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int \frac{x^{2}-1}{x^{3}-3 x+1} d x $$

Step-by-Step Solution

Answer

\( \int \frac{x^2-1}{x^3-3x+1} dx = \frac{1}{3} \ln |x^3 - 3x + 1| + C \).

Step 1: Identify the substitution

We want to simplify the integral by using a substitution. Look at the denominator: \(x^3 - 3x + 1\). Let's substitute \(u = x^3 - 3x + 1\).

Step 2: Differentiate to find du

Differentiate \(u = x^3 - 3x + 1\) with respect to \(x\). We get \( \frac{du}{dx} = 3x^2 - 3 \). So, \( du = (3x^2 - 3)dx \) or \( du = 3(x^2-1)dx \).

Step 3: Solve for dx in terms of du

Given \( du = 3(x^2-1)dx \), we can rearrange to find \( dx \): \( dx = \frac{du}{3(x^2-1)} \).

Step 4: Rewrite the integral using substitution

Replace \(x^2 - 1\) and \(dx\) in the integral with the expressions in terms of \(u\):\[ \int \frac{x^2-1}{x^3-3x+1} dx = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3}\int \frac{1}{u} du \]

Step 5: Integrate with respect to u

The integral \( \frac{1}{3}\int \frac{1}{u} du \) is a basic integral. Integrating, we get \( \frac{1}{3} \ln |u| + C \).

Step 6: Substitute back for x

Since \(u = x^3 - 3x + 1\), substitute back in for \(u\): \( \frac{1}{3} \ln |x^3 - 3x + 1| + C \).

Key Concepts

Substitution Method

The substitution method is a powerful technique to evaluate integrals that might seem challenging at first glance. It simplifies an indefinite integral into a more familiar form, often transforming the original variable into another, allowing for easier integration. In the given exercise, we used substitution to tackle the integral \( \int \frac{x^{2}-1}{x^{3}-3x+1} dx \). The primary step here was noticing that the denominator \( x^3 - 3x + 1 \) can be represented as a single function \( u \). The substitution \( u = x^3 - 3x + 1 \) redefines the integral in terms of \( u \), making the expression easier to handle.To apply this method, always:

• Identify a part of the integral that when substituted, turns the integral into a simpler form.
• Find the derivative of the substitution to express "\( dx \)" in terms of "\( du \)".
• Rewrite the entire integral in terms of \( u \), replacing all \( x \) expressions appropriately.

Substitution is especially beneficial when part of the integral's structure presents itself as the derivative of another part, making the simplification straightforward.

Integration Techniques

Integration techniques are various methods used to find the integral of a function. Choosing the right technique often depends on the structure and components of the function being integrated. In this problem, we employ the substitution method, a common and effective technique.After performing the substitution \( u = x^3 - 3x + 1 \), we transformed our integral into \( \int \frac{1}{u} \cdot \frac{du}{3} \). This new form is simpler because \( \int \frac{1}{u} \, du \) is a standard integral that can be easily integrated using the fundamental rule:

1. The integral \( \int \frac{1}{u} \, du = \ln |u| + C \) is a basic result from calculus.
2. Constants can be factored out, making the computation of the integral straightforward. For example, \( \frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln |u| + C \).

This demonstrates how recognizing standard integrals and applying proper substitutions can turn a complex problem into an elementary one.

Differentiation

Differentiation is the process of finding the derivative, or the rate of change, of a function. It plays a crucial role in solving integrals through substitution.In our exercise, after setting \( u = x^3 - 3x + 1 \), we needed to find its derivative with respect to \( x \) to use in our substitution. Differentiation gives us:\[ \frac{du}{dx} = 3x^2 - 3 \]This derivative helps us express the differential \( du \) in terms of \( dx \), allowing us to change variables inside the integral:\[ du = (3x^2 - 3) dx = 3(x^2 - 1) dx \]Solving for \( dx \), we use this expression to replace "\( dx \)" in the original integral, leading to:\[ dx = \frac{du}{3(x^2-1)} \]Thus, differentiation eases the substitution by providing a critical link to seamlessly transform the variables. This is why understanding derivatives is vital in integration, as it not only finds slopes but also facilitates complex operations like substitutions in integrals.