Problem 23 - Thomas Calculus

Question

Is either of the following equations correct? Give reasons for your answers. a. \(x \int \frac{1}{x} d x=x \ln |x|+C\) b. \(x \int \frac{1}{x} d x=x \ln |x|+C x\)

Step-by-Step Solution

Answer

Equation a is correct; equation b is incorrect due to the extra term added.

Step 1: Understanding the Given Integral

We start with the integral \( \int \frac{1}{x} \, dx \). This integral, based on standard calculus rules, results in the natural logarithm, \( \ln |x| + C \), where \( C \) is the constant of integration. Thus, \( \int \frac{1}{x} \, dx = \ln |x| + C \).

Step 2: Substitute Back into Equation a

Insert the solution from Step 1 into the equation from part a: \( x \cdot \ln |x| + C \). Compare this result to the original expression, which is \( x \ln |x| + C \). The expression on both sides is the same, so the equation a is correct.

Step 3: Analyze Equation b

For part b, substitute the same integral \( \ln |x| + C \) into the expression given: \( x \cdot \ln |x| + C x \). The original result is \( x \ln |x| + C \), but in this case, an extra term \( C x \) is added, which changes the structure of the expression.

Step 4: Conclusion for Equation b

Since the equation in part b introduces an additional term \( C x \), it does not simplify or equate exactly to \( x \ln |x| + C \). Therefore, the equality does not hold true.

Key Concepts

Integration Techniques

When faced with calculus problems, especially those involving integration, it's pivotal to use effective techniques to find solutions. In the case of the integral \( \int \frac{1}{x} \, dx \), the goal is to determine a function whose derivative is \( \frac{1}{x} \). This specific integral is fundamental in calculus and results in the natural logarithm.
To solve integrals, you can use several techniques:

• **Substitution Method**: Useful when the integrand is a function of a function, helping simplify it into a basic form.
• **By Parts**: Applied to integrals of products of functions, using the formula \( \int u \, dv = uv - \int v \, du \).
• **Partial Fractions**: Empowers you to turn complex rational functions into simpler fractions that are easier to integrate.

In our context, recognizing that the integral \( \int \frac{1}{x} \, dx \) simplifies to \( \ln |x| + C \) is key. This forms the basis for evaluating expressions and checking if they hold true, such as in the given equations.

Natural Logarithm

A natural logarithm is a logarithm to the base \(e\), where \(e\) is approximately 2.71828. It is often used in calculus due to its natural properties and connection to the concept of integration.
The natural logarithm, denoted as \( \ln(x) \), has several essential characteristics:

• **Derivative**: The derivative of \( \ln|x| \) with respect to \( x \) is \( \frac{1}{x} \).
• **Range and Domain**: The domain of \( \ln(x) \) is \( x > 0 \), but \( \ln |x| \) can be extended to all \( x eq 0 \), accounting for both positive and negative values.
• **Integral Representation**: The integral \( \int \frac{1}{x} \, dx = \ln |x| + C \) highlights its use in integration techniques.

In our exercise, recognizing the result of the integral as \( \ln |x| + C \) is crucial, as it must precisely match the terms in the equations being evaluated.

Constant of Integration

When performing indefinite integrals, a constant of integration \( C \) is added. This constant accounts for all possible vertical shifts of the antiderivative function.
Here's why the constant is important:

• **Infinite Solutions**: Since differentiation removes constant values, the original function could have had any constant added, hence \( C \).
• **Uniqueness and General Solutions**: Adding \( C \) ensures that all potential solutions are considered, making the solution general rather than specific.
• **Calculation Consistency**: In problems, consistency in including \( C \) is critical to matching integrals to their original expressions accurately. Missing or misunderstanding \( C \) can result in incorrect comparisons, as shown in part b of the exercise.

In the exercise, recognizing the role of the constant \( C \) clarifies why equation a holds true and b does not, since any additional term such as \( Cx \) would disrupt the equation's balance.