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Magazine Chapter 7: Problem 22
Evaluate the integral. \( \displaystyle \int \frac{\ln x}{x \sqrt{1 + (\ln x)^2}}\ dx \)
Short Answer
Expert verified
The integral evaluates to \( \sqrt{1 + (\ln x)^2} + C \).
Step by step solution
01
Recognize substitution
Notice the term \((\ln x)\) appears multiple times. Let us evaluate the integral through substitution. Set \(u = \ln x\) because its derivative, \(du = \frac{1}{x}dx\), seems to align well with the given integral.
02
Perform substitution
Given \(u = \ln x\), we have \(du = \frac{1}{x} dx\). Substitute in the integral, replacing \(\ln x\) with \(u\) and \(\frac{1}{x} dx\) with \(du\). This gives us the integral \(\int \frac{u}{\sqrt{1 + u^2}}\ du\).
03
Setup new substitution
Now consider the integral \(\int \frac{u}{\sqrt{1+u^2}}\ du\). Use a substitution technique suited for this, like setting \(v = 1 + u^2\), thus \(dv = 2u du\).
04
Simplify with substitution
From \(dv = 2u du\), we have \(u du = \frac{1}{2} dv\). Substitute back into the integral: \(\int \frac{u}{\sqrt{1+u^2}}\ du = \frac{1}{2} \int \frac{1}{\sqrt{v}} dv\), since \(v = 1 + u^2\).
05
Perform the integration
The integral \(\int \frac{1}{\sqrt{v}} dv\) is a standard integral. It evaluates to \(2\sqrt{v} + C\). Substitute back to get \(\frac{1}{2}(2\sqrt{1+u^2}) + C = \sqrt{1+u^2} + C\).
06
Undo substitution
Substitute back \(u = \ln x\) to return to the original variable. Thus \(\sqrt{1+u^2} = \sqrt{1+(\ln x)^2}\). Therefore, the integral becomes \(\sqrt{1+(\ln x)^2} + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used in integration. It simplifies complex integrals using a change of variable. In this method, we substitute part of the integral with a new variable. This alteration often makes the integral easier to solve.
Let's break it down:
Let's break it down:
- Identify a part of the integral that can be substituted. It usually involves repetitive terms or functions with recognizable derivatives.
- Assign a new variable to this repeated term, making sure its derivative is present in the integral.
- Replace the identified part and adjust the remaining differential elements to reflect the new variable.
Natural Logarithm
The natural logarithm, often denoted as \( \ln x \), is a logarithm based on the constant \( e \), where \( e \approx 2.71828 \). It is the inverse operation of exponentiation involving \( e \).
Why it matters:
Why it matters:
- The natural logarithm is prevalent in various mathematical contexts, especially calculus.
- It has unique properties that make it useful for integration, differentiation, and solving equations.
- When differentiating, the derivative of \( \ln x \) is \( \frac{1}{x} \), which is extremely helpful in substitution, as seen in our problem.
Standard Integral
Standard integrals are pre-calculated integrations of common functions. They simplify the process of solving integrals by providing us with previously derived solutions for certain forms.
Advantages of using standard integrals:
Advantages of using standard integrals:
- Time-efficient, as they eliminate the need to perform lengthy calculations.
- Useful as benchmarks for both educators and students when solving complex integrals.
- Act as building blocks to solve more intricate integral problems and equations.
Definite and Indefinite Integrals
Integrals are broadly categorized into definite and indefinite integrals, each serving distinct purposes.
- Indefinite Integrals: Represent the antiderivative of a function. They include a constant of integration, \( C \), as they represent a family of functions.
- Definite Integrals: Evaluate the area under a curve between two points. They result in a specific numerical value and do not contain the integration constant \( C \).
