Problem 34 - Calculus for Biology and Medicine

Question

Use substitution to evaluate the indefinite integrals. $$ \int \frac{d x}{(x-3) \ln (x-3)} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |\ln(x - 3)| + C \).

Step 1: Choose a Substitution Variable

To simplify the integral, first choose a substitution variable. Let \( u = x - 3 \). This substitution will simplify the expression inside the integrand.

Step 2: Differentiate the Substitution

Find the derivative of the substitution with respect to \( x \): \( du = dx \). This transforms the integral's differential part.

Step 3: Rewrite the Integral

Substitute \( u = x - 3 \) and \( du = dx \) into the integral: \[ \int \frac{du}{u \ln u} \].

Step 4: Use a Second Substitution

Let \( v = \ln u \), hence \( dv = \frac{1}{u} du \). Then, the integral becomes \( \int \frac{du}{u \ln u} = \int \frac{dv}{v} \).

Step 5: Integrate

Integrate \( \int \frac{1}{v} dv \), which yields \( \ln |v| + C \), where \( C \) is the constant of integration.

Step 6: Back-Substitute the Variables

Replace \( v \) and \( u \) with their original expressions: \( v = \ln u \) and \( u = x - 3 \). Therefore, the solution is \( \ln |\ln(x - 3)| + C \).

Key Concepts

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus. They represent the general form of antiderivatives of a function. When you take an indefinite integral, you are essentially finding the most general form of a function whose derivative is the integrand.
An indefinite integral is expressed in the form \( \int f(x) \, dx \), where \( f(x) \) is the function being integrated. It also includes a constant of integration, usually denoted by \( C \), because differentiation of a constant yields zero, meaning there could be infinitely many antiderivatives.

• The indefinite integral provides a family of functions rather than a single solution.
• The result includes the integration constant \( C \) to account for all possible vertical shifts of the antiderivative graph.
• It’s important in applications like finding original functions from known derivatives or solving differential equations.

In the provided exercise, we find the indefinite integral of \( \frac{1}{(x-3) \ln(x-3)} \) using substitution, which simplifies the process of integration considerably.

Integration by Substitution

Integration by substitution is a technique used to simplify difficult integrals by changing variables. It resembles the reverse of the chain rule used in differentiation. The goal is to transform a complex integral into a simpler one. This is achieved by choosing a new variable \( u \) to replace a function of \( x \) in the integrand.

• The choice of substitution \( u = x - 3 \) simplifies the structure of the integral, as seen in the given exercise.
• After selecting the substitution, differentiate \( u \) to find \( du \), which allows the transformation of the differential \( dx \) into \( du \).
• Rewrite the original integral in terms of \( u \) and \( du \) to make integration simpler.

In the exercise, a second substitution \( v = \ln u \) further simplifies the problem, transforming a challenging integral into a familiar form. Thus, substitution can be a powerful method, especially when dealing with nested functions or composite functions.

Calculus Techniques

Calculus involves a rich collection of techniques for solving integrals and other mathematical problems. The substitution method used in the exercise is just one of the many techniques available. Here are some key insights into similar calculus techniques:

• By Parts Method: Useful when the substitution doesn't simplify the integral, involving product of functions (\( \int u dv = uv - \int v du \)).
• Partial Fractions: Decomposes complex rational expressions into simpler ones, aiding integration.
• Trigonometric Integrals: Useful when dealing with products of sine and cosine functions, often utilizing trigonometric identities for simplification.

Each technique serves as a tool to tackle the form and complexity of the problem at hand. Mastering these techniques helps in addressing a variety of problems, like the one solved through substitution in the exercise, ultimately building a robust understanding of integral calculus.