Problem 80 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{1}{x-3} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the function to integrate

The function inside the integral is \( \frac{1}{x-3} \). This suggests that the integral may involve the natural logarithm, as the derivative of \( \ln |x| \) is \( \frac{1}{x} \).

Step 2: Use the substitution method

Set \( u = x - 3 \), so that \( du = dx \). This substitution simplifies the integral to \( \int \frac{1}{u} \, du \).

Step 3: Integrate using the natural logarithm

The integral \( \int \frac{1}{u} \, du \) equals \( \ln |u| + C \), where \( C \) is the constant of integration. Thus, we have \( \ln |x - 3| + C \) after substituting back \( u = x - 3 \).

Step 4: Write the final result

Therefore, the indefinite integral of the function is \( \int \frac{1}{x-3} \ dx = \ln |x-3| + C \).

Key Concepts

Understanding Indefinite Integrals

In calculus, integrals are a fundamental concept used to find areas under curves and reverse differentiation. An indefinite integral is often referred to as the antiderivative. It represents a family of functions whose derivative gives the original function. For example, if we start with a function like \( f(x) = \frac{1}{x-3} \), finding its indefinite integral would mean recovering a function whose derivative would return \( f(x) \). These integrals include a constant of integration, denoted as \( C \), to account for any constant that might have been nullified through differentiation. This is very important since differentiation of a constant is zero and doesn't change the shape of the derivative's curve. When working through indefinite integrals, we often rely on memorized rules and techniques from calculus, which make the computation straightforward.

The Substitution Method in Calculus

The substitution method is a key technique in calculus for simplifying integrals. It often provides a straightforward path to the solution by changing variables, which makes challenging integrals more manageable. The primary goal of substitution is to rewrite a complex integral into a simpler form that we already know how to integrate. For instance, consider the integral \( \int \frac{1}{x-3} \, dx \). By letting \( u = x - 3 \), we simplify the function to \( \int \frac{1}{u} \, du \), which is a form that we can integrate easily.

• First, choose a substitution. Often, this means setting \( u \) equal to an expression within the integral.
• Calculate \( du \) to replace \( dx \), ensuring the entire integral is expressed in terms of \( u \).
• Perform the integration in the transformed form.
• Finally, substitute back the original variable to complete the process.

This method showcases the flexibility of calculus, allowing complex problems to be broken down into familiar, simpler tasks.

Exploring the Natural Logarithm in Integration

Natural logarithms are an essential part of integration techniques, particularly when dealing with fractions that have linear denominators. The natural logarithm, denoted as \( \ln \), is the inverse function of the exponential function \( e^x \). In integration, this function often appears when integrating expressions of the form \( \frac{1}{x} \). A critical rule to remember is that the integral \( \int \frac{1}{u} \, du \) equals \( \ln |u| + C \), where \( C \) is the constant of integration. In our example \( \int \frac{1}{x-3} \, dx \), by using substitution \( u = x - 3 \), it transforms to \( \int \frac{1}{u} \, du \), which evaluates to \( \ln |u| + C \), or written in terms of \( x \), \( \ln |x-3| + C \).

• The appearance of the absolute value is crucial since the domain of \( \ln(x) \) is positive real numbers.
• The constant \( C \) represents the family of antiderivatives, reflecting the arbitrary constant lost through differentiation.

Understanding these principles helps to extend the use of natural logarithms in solving calculus problems effectively.