Chapter 28: Problem 21
Integrate each of the given functions. $$\int \frac{d r}{r \ln r}$$
Short Answer
Expert verified
The integral is \( \ln | \ln r | + C \).
Step by step solution
01
Identify the Substitution
Recognize that the function inside the integral has the form suitable for a logarithmic substitution. We can choose a substitution to simplify the integration process.
02
Perform the Substitution
Let \( u = \ln r \). Then, \( du = \frac{1}{r} dr \). This implies that \( dr = r \, du \), and since \( u = \ln r \), it follows that \( r = e^u \) (by the definition of the natural logarithm). Substitute into the integral.
03
Substitute and Simplify
Substitute \( r \) and \( du \) into the integral: \[ \int \frac{dr}{r \ln r} = \int \frac{1}{u} \cdot r \, du = \int \frac{1}{u} \, e^u \, du \]. Simplify this using the substitution relationships.
04
Recognize the New Form
Realize that the integral \( \int e^u \, du \) is a standard form. Since \( \frac{1}{u} \, du = du \, \) this simplifies our problem significantly.
05
Integrate
Now integrate the expression \( \int e^u \, du = e^u + C \), where \( C \) is the constant of integration.
06
Substitute Back
Return to the original variable by substituting back \( u = \ln r \). The solution to the integral is \( e^{\ln r} + C = r + C \), since \( e^{\ln r} = r \).
07
Final Answer
The integral \( \int \frac{dr}{r \ln r} = \ln | \ln r | + C \). Note: The constant \( C \) accounts for the indefinite integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Substitution
Logarithmic substitution is a clever technique used in integration, particularly when dealing with functions involving logarithms. It helps simplify the problem by transforming it into a more manageable form.
The basic idea is to replace a complicated part of the function with a new variable. In our integral \( \int \frac{dr}{r \ln r} \), identifying the right substitution is key. Here, making the substitution \( u = \ln r \) replaces the pesky logarithm with \( u \), simplifying the process considerably.
Once you've chosen \( u = \ln r \), the differential \( du \) becomes \( \frac{1}{r} dr \). This conversion is crucial because it helps rewrite the integral in terms of \( u \) and \( du \), making it easier to solve. The original expression then transforms into \( \int \frac{1}{u} e^u \, du \), which is much simpler to integrate.
The basic idea is to replace a complicated part of the function with a new variable. In our integral \( \int \frac{dr}{r \ln r} \), identifying the right substitution is key. Here, making the substitution \( u = \ln r \) replaces the pesky logarithm with \( u \), simplifying the process considerably.
Once you've chosen \( u = \ln r \), the differential \( du \) becomes \( \frac{1}{r} dr \). This conversion is crucial because it helps rewrite the integral in terms of \( u \) and \( du \), making it easier to solve. The original expression then transforms into \( \int \frac{1}{u} e^u \, du \), which is much simpler to integrate.
- Choose \( u \) to make the integral easier.
- Ensure you can express the original variable in terms of \( u \) and \( du \).
- Convert the integral fully to the new variable before solving.
Indefinite Integrals
Indefinite integrals are a type of integral that represent a family of functions. Unlike definite integrals, they do not have specific upper and lower limits for integration. Instead, they include a constant \( C \), which accounts for any constant difference in the solutions.
When you solve an indefinite integral, such as \( \int e^u \, du \), your result is a general form, like \( e^u + C \). This constant \( C \) means that there are multiple functions whose derivative could give the original integrand.
In our exercise, solving the integral results in \( \int \frac{dr}{r \ln r} \) leading to \( \ln | \ln r | + C \). This symbol \( C \) is crucial because without it, the solution would not capture all possible forms of the original function.
When you solve an indefinite integral, such as \( \int e^u \, du \), your result is a general form, like \( e^u + C \). This constant \( C \) means that there are multiple functions whose derivative could give the original integrand.
In our exercise, solving the integral results in \( \int \frac{dr}{r \ln r} \) leading to \( \ln | \ln r | + C \). This symbol \( C \) is crucial because without it, the solution would not capture all possible forms of the original function.
- Indefinite integrals solve for a function without specific boundaries.
- The constant \( C \) allows for an infinite number of solutions.
- A correct indefinite integral accounts for every possible function with the given derivative.
Change of Variables
Change of variables, also known as substitution, is a fundamental technique in integration. The goal here is to replace a complex or difficult part of the integrand with a new variable that simplifies the integration process.
This technique was vital in the exercise \( \int \frac{dr}{r \ln r} \). By letting \( u = \ln r \), we changed the variable from \( r \) to \( u \). This altered the integral into \( \int \frac{1}{u} e^u \, du \), greatly simplifying the task.
The process involves a few crucial steps:
This technique was vital in the exercise \( \int \frac{dr}{r \ln r} \). By letting \( u = \ln r \), we changed the variable from \( r \) to \( u \). This altered the integral into \( \int \frac{1}{u} e^u \, du \), greatly simplifying the task.
The process involves a few crucial steps:
- Identify a substitution that will make the integral simpler.
- Convert the original variable and differential into the new variable.
- Translate the entire integral into terms of the new variable.
- After integration, return to the original variable by reversing the substitution.