Chapter 11: Problem 939
\(\int\left[(\log \mathrm{x}) /(1+\log \mathrm{x})^{2}\right] \mathrm{d} \mathrm{x}=\underline{ }+\mathrm{c}\) (a) \([\mathrm{x} /(1+\log \mathrm{x})]\) (b) \(x(1+\log x)\) (c) \([\mathrm{x} /(\log \mathrm{x})]\) (d) \(x \log x+x^{-1}\)
Short Answer
Expert verified
(a) $[\frac{x}{1+\log x}]+c$
Step by step solution
01
Define the Substitution Variable
Let \(u = 1 + \log x\). The goal is to rewrite the integral in terms of u, allowing us to integrate the function more easily.
02
Find the Derivative of the Substitution Variable
Differentiate u with respect to x: \(\frac{du}{dx} = \frac{d}{dx}(1+\log x) = \frac{1}{x}\) (because the derivative of 1 is 0 and the derivative of \(\log x\) is \(\frac{1}{x}\)).
03
Solve for dx
Isolate dx by multiplying both sides by dx and dividing both sides by \(\frac{1}{x}\): \(dx = x\ du\)
04
Replace the Original Variables with Substitution Variable
Substitute the variables according to the substitution variable definition (u) and the calculated dx: \(\int\frac{\log x}{(1 + \log x)^2} dx = \int\frac{u - 1}{u^2} \cdot x\ du\)
05
Simplify the Integral
Simplify the integral: \(\int\frac{u - 1}{u^2} \cdot x\ du = \int\frac{u}{u^2} du - \int\frac{1}{u^2} du\)
06
Integrate the Separated Functions
Integrate both functions: \(\int\frac{u}{u^2} du = \int u^{-1} du\) and \(\int\frac{1}{u^2} du = \int u^{-2} du\)
The integral of the first function is: \(\int u^{-1} du = \log|u|\)
The integral of the second function is: \(\int u^{-2} du = \frac{u^{-1}}{-1}\)
Combine both integrals, and don't forget the constant (c): \(\log|u| - \frac{1}{u} + c\)
07
Substitute the Original Variable Back
Replace u with the original expression that was defined in Step 1 (u = 1 + \log x):
\(\log|1 + \log x| - \frac{1}{(1 + \log x)} + c\)
The solution is (a): \([\frac{x}{1+\log x}]+c\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indefinite Integral
Calculating an indefinite integral, also known as antiderivative, involves finding a function that represents all the antiderivatives of a given function. The result includes an arbitrary constant, often denoted as 'c', to represent the family of all possible solutions. The notation for an indefinite integral is shown as \( \int f(x) \, dx \), where \( f(x) \) is the integrand and \( dx \) indicates that the integration is with respect to the variable \( x \).
For example, the indefinite integral of \( x^2 \) is \( \frac{1}{3}x^3 + c \) because the derivative of \( \frac{1}{3}x^3 \) is \( x^2 \). This process is fundamental in calculus, as it allows us to determine quantities like area under a curve and the antiderivative of velocity, which is position.
For example, the indefinite integral of \( x^2 \) is \( \frac{1}{3}x^3 + c \) because the derivative of \( \frac{1}{3}x^3 \) is \( x^2 \). This process is fundamental in calculus, as it allows us to determine quantities like area under a curve and the antiderivative of velocity, which is position.
Logarithmic Integration
Logarithmic integration involves integrating functions with logarithms. A key fact is that the integral of \( 1/x \) is \( \log|x| + c \), where \( c \) is the integration constant. When dealing with more complex logarithmic expressions, substitution usually simplifies the integral into a form where this relationship can be applied. This technique is especially useful when the argument of a logarithmic function appears in the numerator of a fraction.
For instance, the indefinite integral of \( (\log x)/x \) can directly be integrated to \( (1/2)\log^2|x| + c \) by recognizing the presence of the function and its derivative. This reflects the inseparable relationship between logarithmic functions and their integrals in calculus.
For instance, the indefinite integral of \( (\log x)/x \) can directly be integrated to \( (1/2)\log^2|x| + c \) by recognizing the presence of the function and its derivative. This reflects the inseparable relationship between logarithmic functions and their integrals in calculus.
U-Substitution
U-Substitution is a technique designed to simplify integration by changing the variable of integration into something more manageable. It's particularly useful when dealing with compositions of functions, like a logarithm under a power, as seen in the given exercise. The general idea is to let \( u \) equal a part of the integral that makes the derivative simpler, often to cancel out other parts of the integrand when \( dx \) is substituted.
To execute u-substitution, we differentiate \( u \) with respect to \( x \), solve for \( dx \), and substitute all instances of \( x \) with \( u \) in the integrand. After integration, we 'back-substitute' \( u \) with the original variable's terms. This method can significantly reduce the complexity of an integration problem and is a staple in any calculus toolkit.
To execute u-substitution, we differentiate \( u \) with respect to \( x \), solve for \( dx \), and substitute all instances of \( x \) with \( u \) in the integrand. After integration, we 'back-substitute' \( u \) with the original variable's terms. This method can significantly reduce the complexity of an integration problem and is a staple in any calculus toolkit.
Integration Techniques
A variety of integration techniques exist to tackle different kinds of integrals. Besides u-substitution, there are methods such as integration by parts, partial fractions, and trigonometric integration, among others. Knowing when to apply a specific technique comes with practice and understanding the form of the function you are integrating.
For instance, integration by parts is suitable for products of functions, and partial fractions work well with rational functions that can be decomposed. Trigonometric integration helps with functions involving sine and cosine functions. Mastering these techniques allows you to integrate a wide array of functions, enabling you to solve more complex calculus problems with confidence.
For instance, integration by parts is suitable for products of functions, and partial fractions work well with rational functions that can be decomposed. Trigonometric integration helps with functions involving sine and cosine functions. Mastering these techniques allows you to integrate a wide array of functions, enabling you to solve more complex calculus problems with confidence.