Chapter 7: Problem 110
Evaluate the integrals. \begin{equation}\int \frac{d x}{x\left(\log _{8} x\right)^{2}}\end{equation}
Short Answer
Expert verified
The integral evaluates to \(-\frac{1}{\log_8 x} + C\).
Step by step solution
01
Recognize Type of Integral
Notice that the integral \( \int \frac{d x}{x(\log_{8} x)^{2}} \) is a case where it is beneficial to use substitution involving the logarithm term.
02
Perform Substitution
Let \( u = \log_{8} x \). Then, noticing that \( \log_{8} x = \frac{\ln x}{\ln 8} \), differentiate both sides with respect to \( x \): \( \frac{d}{dx}(\log_{8} x) = \frac{1}{x \ln 8} \). Thus \( du = \frac{1}{x \ln 8} dx \), giving \( dx = x \ln 8 \, du \).
03
Substitute and Simplify the Integral
Substitute \( u \) and \( dx \) into the integral: \[ \int \frac{x \ln 8 \, du}{x u^2} = \int \frac{\ln 8}{u^2} \, du = \ln 8 \int u^{-2} \, du. \]
04
Integrate Using Power Rule
Integrate \( \ln 8 \int u^{-2} du \) using the power rule for integrals. The result is: \[ \ln 8 \cdot \left( -u^{-1} \right) = -\frac{\ln 8}{u}. \]
05
Substitute Back Original Variable
Replace \( u \) with \( \log_{8} x \) to revert back to the original variable:\[ -\frac{\ln 8}{\log_{8} x} + C = -\frac{1}{\log_{8} x / \ln 8} + C = -\frac{\ln 8}{\log_8 x} + C. \]
06
Simplification
Simplify the expression, keeping \( C \) as the integration constant:\[ -\frac{\ln 8}{\log_8 x} + C. \]
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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 method used to evaluate integrals that involve logarithmic functions. In many calculus problems, integrals feature expressions like \( \log_{b} x \). These types of integrals can sometimes be simplified further using substitutions, allowing for easier integration.
For example, when faced with \( \int \frac{1}{x(\log_{8} x)^2} \, dx \), we identify the presence of the logarithmic expression \( \log_{8} x \). By setting this portion equal to a new variable, say \( u \), we can streamline the process. Later, through integration by substitution, we change the variables and simplify the integral into a more manageable form. This highlights how identifying a logarithmic part can lead to a simpler integration path.
For example, when faced with \( \int \frac{1}{x(\log_{8} x)^2} \, dx \), we identify the presence of the logarithmic expression \( \log_{8} x \). By setting this portion equal to a new variable, say \( u \), we can streamline the process. Later, through integration by substitution, we change the variables and simplify the integral into a more manageable form. This highlights how identifying a logarithmic part can lead to a simpler integration path.
U-Substitution
U-substitution is a powerful technique in integration, often allowing us to transform a complex integral into a simpler form. This process involves substituting a portion of the integral with a single variable, typically \( u \).
For the given exercise, we start by defining \( u = \log_{8} x \). This substitution choice is crucial because it helps simplify the integral where \( du \) corresponds to the derivative of \( \log_{8} x \), which is \( \frac{1}{x \ln 8} \, dx \).
Thus, we find that:
For the given exercise, we start by defining \( u = \log_{8} x \). This substitution choice is crucial because it helps simplify the integral where \( du \) corresponds to the derivative of \( \log_{8} x \), which is \( \frac{1}{x \ln 8} \, dx \).
Thus, we find that:
- \( du = \frac{1}{x \ln 8} \, dx \)
- \( dx = x \ln 8 \, du \)
Power Rule of Integration
The power rule is a fundamental aspect of integration. This rule is applicable for integrating functions of the form \( x^n \). It states that \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] provided \( n eq -1 \).
In our exercise, after substitution, the integral \( \ln 8 \int u^{-2} \, du \) arises, which fits the power rule. We substitute and simplify using the rule:
- Integrate \( u^{-2} \), which gives \( -u^{-1} \).
- Hence, the integral becomes \( \ln 8 \cdot (-u^{-1}) \). This step involves treating \( \ln 8 \) as a constant multiplier and solving the simple antiderivative. Applying the power rule turns what appears complex into a straightforward process.
In our exercise, after substitution, the integral \( \ln 8 \int u^{-2} \, du \) arises, which fits the power rule. We substitute and simplify using the rule:
- Integrate \( u^{-2} \), which gives \( -u^{-1} \).
- Hence, the integral becomes \( \ln 8 \cdot (-u^{-1}) \). This step involves treating \( \ln 8 \) as a constant multiplier and solving the simple antiderivative. Applying the power rule turns what appears complex into a straightforward process.
Integration by Substitution
Integration by substitution is a method that can effectively tackle integrals resembling a chain rule derivative. It's particularly useful for integrals where a direct application of basic rules isn't apparent.
In our example, substituting \( u = \log_{8} x \) simplifies the integral into a form that can be managed with the power rule. This strategy, known as "u-substitution," mirrors techniques in differential calculus but for antiderivatives.
Consider these steps during our substitution:
In our example, substituting \( u = \log_{8} x \) simplifies the integral into a form that can be managed with the power rule. This strategy, known as "u-substitution," mirrors techniques in differential calculus but for antiderivatives.
Consider these steps during our substitution:
- Identify terms in the integral that seem to call for differentiation (e.g., \( \log_{8} x \)).
- Substitute these terms with a simple variable \( u \) and differentiate to find \( du \).