Chapter 8: Problem 11
Evaluate the following integrals. $$\int \frac{\ln 2 x}{x} d x$$
Short Answer
Expert verified
Question: Evaluate the integral: $$\int \frac{\ln 2x}{x} dx$$
Answer: $$\int \frac{\ln 2x}{x} dx = \ln(2x)\cdot \ln|x| -\ln|x|\cdot\ln|x|+C$$
Step by step solution
01
1. Choose u and dv
Let \(u=\ln(2x)\) and \(dv=\frac{1}{x}dx\).
02
2. Find du
Differentiate \(u\) with respect to \(x\):
$$du = \frac{d}{dx}(\ln(2x)) dx$$
Using chain rule,
$$du = \frac{1}{2x}(2) dx = \frac{1}{x}dx$$
03
3. Find v
Integrate \(dv\):
$$v = \int \frac{1}{x} dx$$
Using the fact that \(\int \frac{1}{x} dx = \ln|x| + C\):
$$v = \ln|x|$$
04
4. Apply integration by parts formula
Using the integration by parts formula:
$$\int u dv = uv - \int v du$$
Substitute our values for \(u\), \(v\), and \(du\):
$$\int \frac{\ln 2x}{x} dx = \ln(2x) \cdot \ln|x| - \int \ln|x| \cdot \frac{1}{x} dx$$
05
5. Integrate remaining term
Now we need to integrate the remaining term:
$$\int \ln|x| \cdot \frac{1}{x} dx$$
We can apply the integration by parts formula once more with \(u = \ln|x|\) and \(dv = \frac{1}{x}dx\). Following the same steps, we have:
$$du = \frac{1}{x}dx$$
$$v = \ln|x|$$
Therefore,
$$\int \ln|x| \cdot \frac{1}{x} dx = \ln|x|\cdot\ln|x| - \int \ln|x| \cdot \frac{1}{x} dx$$
06
6. Solve the integral
Now, we can solve the integral using the result from step 5:
$$\int \frac{\ln 2x}{x} dx = \ln(2x) \cdot \ln|x| - \left(\ln|x|\cdot\ln|x| - \int \ln|x| \cdot \frac{1}{x} dx\right)$$
Simplifying the equation, we get:
$$\int \frac{\ln 2x}{x} dx = \ln(2x) \cdot \ln|x| - \ln|x|\cdot\ln|x| + \int \ln|x| \cdot \frac{1}{x} dx$$
We can now cancel the integral on both sides of the equation:
$$\int \frac{\ln 2x}{x} dx =\ln(2x)\cdot \ln|x| -\ln|x|\cdot\ln|x|+C$$
This is our final answer:
$$\int \frac{\ln 2x}{x} dx = \ln(2x)\cdot \ln|x| -\ln|x|\cdot\ln|x|+C$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Integration
Logarithmic integration is a technique used in calculus to evaluate integrals involving logarithmic functions. This often requires the method known as 'integration by parts', a process that breaks down a complex integral into simpler parts. In this specific exercise, we have dealt with the integral \( \int \frac{\ln 2x}{x} \, dx \). Here, the presence of the natural logarithm \( \ln(2x) \), combined with \( \frac{1}{x} \), makes it a candidate for integration by parts.
When we use integration by parts, we typically assign \( u = \ln(2x) \) and \( dv = \frac{1}{x} \, dx \). After differentiating \( u \) and integrating \( dv \), we apply the formula:
When we use integration by parts, we typically assign \( u = \ln(2x) \) and \( dv = \frac{1}{x} \, dx \). After differentiating \( u \) and integrating \( dv \), we apply the formula:
- \( \int u \, dv = u v - \int v \, du \)
Definite Integrals
Definite integrals involve evaluating the integral of a function within a given interval, providing the total accumulation of quantities. Unlike indefinite integrals, which represent a family of functions, definite integrals compute an exact numerical value.
In our exercise, while we began with an indefinite integral \( \int \frac{\ln 2x}{x} \, dx \), understanding definite integrals is crucial for similar problems involving specific limits. Suppose we were to calculate the accumulation between \( a \) and \( b \), we would represent it as:
In our exercise, while we began with an indefinite integral \( \int \frac{\ln 2x}{x} \, dx \), understanding definite integrals is crucial for similar problems involving specific limits. Suppose we were to calculate the accumulation between \( a \) and \( b \), we would represent it as:
- \[ \int_{a}^{b} \frac{\ln 2x}{x} \, dx \]
- \[ F(b) - F(a) \]
Calculus Problem Solving
Problem-solving is at the heart of calculus and involves applying multiple calculus concepts to find solutions. In our exercise's context, several concepts are entwined: logarithmic functions, integration by parts, and simplifying expressions. To successfully solve calculus problems, especially those like \( \int \frac{\ln 2x}{x} \, dx \), you need a structured approach:
- Identify functions that can be simplified or separated, like distinguishing parts for integration by parts.
- Apply calculus techniques: Understand when to use techniques like the chain rule or the integration by parts formula.
- Solve systematically: Work step-by-step as shown in the solution, ensuring every differentiation and integration step is correct.
- Constantly check: Simplify your final expression and verify calculations.