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Chapter 7: Problem 44

Evaluate the integrals. $$\int_{2}^{3} \frac{2 \log _{2}(x-1)}{x-1} d x$$

Short Answer

Expert verified
The integral evaluates to \( \ln 2 \).

Step by step solution

01

Identify the Integral

The integral we need to evaluate is \( \int_{2}^{3} \frac{2 \log_{2}(x-1)}{x-1} \, dx \). This can be simplified by noticing that \( \frac{2 \log_{2}(x-1)}{x-1} = 2 \cdot \frac{\log_{2}(x-1)}{x-1} \).
02

Change of Base Formula for Logarithm

Recall the change of base formula: \( \log_{a}(b) = \frac{\ln(b)}{\ln(a)} \). Thus, \( \log_{2}(x-1) = \frac{\ln(x-1)}{\ln(2)} \). Substitute this into the integral, yielding: \[ 2 \int_{2}^{3} \frac{\ln(x-1)}{(x-1) \ln(2)} \, dx = \frac{2}{\ln(2)} \int_{2}^{3} \frac{\ln(x-1)}{x-1} \, dx. \]
03

Recognize the Form of a Known Integral

Notice that \( \int \frac{\ln(u)}{u} \, du = \frac{1}{2} \ln^{2} |u| + C \). In this case, \( u = x-1 \), so we have \( \int \frac{\ln(x-1)}{x-1} \, dx = \frac{1}{2} \ln^{2} |x-1| + C \).
04

Integrate and Apply the Limits

Using the antiderivative from Step 3, compute the definite integral: \[ \frac{1}{2} \ln^{2} |x-1| \bigg|_{2}^{3} = \frac{1}{2} \ln^{2} |3-1| - \frac{1}{2} \ln^{2} |2-1|. \] Simplify the expression: \[ \frac{1}{2} \ln^{2} 2 - \frac{1}{2} \ln^{2} 1 = \frac{1}{2} (\ln^{2} 2 - 0) = \frac{1}{2} \ln^{2} 2. \]
05

Include the Constant Factor

Recall the factor \( \frac{2}{\ln(2)} \) from Step 2. Multiply this by the result from Step 4: \[ \frac{2}{\ln(2)} \times \frac{1}{2} \ln^{2} 2 = \ln 2. \] Thus, the evaluated integral is \( \ln 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Integration
In calculus, the integration of logarithmic functions is a process that can initially seem tricky, yet with the right steps, it becomes much easier. In particular, the expression \( \frac{\ln(u)}{u} \) stands out as a commonplace form. When integrating functions that involve logarithms, recognizing these forms is crucial. It provides a shortcut to finding the antiderivative or the integral.
  • The integral \( \int \frac{\ln(u)}{u} \, du \) results in \( \frac{1}{2} \ln^2 |u| + C \), where \( C \) is the constant of integration, a crucial step for indefinite integrals. However, when it comes to definite integrals, we focus on the bounds rather than the constant.
  • In our problem, \( u = x-1 \). By applying \( \int \frac{\ln(x-1)}{x-1} \, dx \) directly, we simplify the process significantly.
Understanding this integral is key for students tackling similar problems, as it often appears in calculus courses. Becoming familiar with logarithmic integration will ensure you are well-prepared to tackle more complex integrals.
Change of Base Formula
The change of base formula is a powerful tool when dealing with logarithms. It allows the conversion of logarithms from one base to another, usually to the natural logarithm (ln), which is often easier to handle due to its prevalence in calculus.
  • The formula is \( \log_{a}(b) = \frac{\ln(b)}{\ln(a)} \). This conversion allows students to transform logarithms into a base that works more seamlessly with integration techniques.
  • In the given problem, changing \( \log_{2}(x-1) \) to \( \frac{\ln(x-1)}{\ln(2)} \) simplifies the integral drastically. Particularly in calculus, where natural logarithms often appear, this change can expose paths to solutions that were previously obscured.
When engaging with problems involving base conversion, keep in mind the benefits of simplifying expressions to their natural logarithmic form. This makes the integration process not only simpler but also aligns it with the forms of integrals commonly taught.
Definite Integral Evaluation
Evaluating definite integrals involves calculating the net area under a curve between two specific points. The definite integral \( \int_{a}^{b} f(x) \, dx \) calculates the accumulation of \( f(x) \) from \( a \) to \( b \). In the final steps of solving a definite integral, utilize the antiderivative then apply the Fundamental Theorem of Calculus. This principle is
  • The antiderivative is calculated, followed by applying the upper and lower limits, which gives the difference of their respective outputs.
  • Here, the calculated antiderivative \( \frac{1}{2} \ln^{2}|x-1| \) is found first. Then, evaluating between 2 and 3 gives the expression \( \frac{1}{2} \ln^{2} (2) - \frac{1}{2} \ln^{2} (1) \). Since \( \ln(1)=0 \), it simplifies to \( \frac{1}{2} \ln^{2}(2) \).
Lastly, multiplying by the external constant derived earlier wraps it all together, simplifying to the neat final result. Definite integrals, once broken down into these steps, become a manageable and systematic process to arrive at a precise solution.

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Most popular questions from this chapter

Verify the integration formulas. $$\int \tanh ^{-1} x d x=x \tanh ^{-1} x+\frac{1}{2} \ln \left(1-x^{2}\right)+C$$

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