Problem 70 - Thomas Calculus

Question

Evaluate the integrals. \(\int \frac{d x}{x\left(\log _{8} x\right)^{2}}\)

Step-by-Step Solution

Answer

The integral is \(-\frac{(\ln 8)^2}{\ln x} + C\)."

Step 1: Variable Substitution

To simplify the integral, use a substitution. Let \( u = \log_{8} x \). Then, we have \( x = 8^u \) and \( dx = 8^u \ln(8) \, du \). Substituting these into the integral yields:\[\int \frac{dx}{x (\log_{8} x)^2} = \int \frac{8^u \ln(8) \, du}{8^u u^2} = \int \frac{\ln(8) \, du}{u^2}\]

Step 2: Integrate Using Power Rule

Now, integrate \( \int \frac{\ln(8)}{u^2} \, du \). We can factor out the constant \( \ln(8) \), giving:\[\ln(8) \int u^{-2} \, du\]The integral of \( u^{-2} \) is \( -u^{-1} \) (power rule for integration). Therefore:\[\ln(8) \left( -\frac{1}{u} \right) + C = -\frac{\ln(8)}{u} + C\]

Step 3: Substitute Back

Replace \( u \) with \( \log_{8} x \) to express the integral in terms of \( x \):\[-\frac{\ln(8)}{\log_{8} x} + C\]

Step 4: Simplify Using Change of Base Formula

Use the change of base formula \( \log_{8} x = \frac{\ln x}{\ln 8} \) to further simplify the expression:\[-\frac{\ln(8)}{\frac{\ln x}{\ln 8}} = -\frac{(\ln(8))^2}{\ln x} + C\]

Key Concepts

Substitution Method

The substitution method is a powerful tool in integral calculus that simplifies complex integrals by transforming them into simpler ones. In this exercise, we used the substitution \( u = \log_{8} x \). This transformation simplifies the integrative process:

• First, express \( x \) in terms of \( u \), giving \( x = 8^u \).
• Calculate \( dx \) in terms of \( du \) to replace it in the integral, so \( dx = 8^u \ln(8) \, du \).
• Substitute \( x \) and \( dx \) in the original integral. This simplifies the complex expression into a simpler form: \( \int \frac{\ln(8) \, du}{u^2} \).

Substitution transforms the integral into a more manageable form, where standard techniques like the power rule can be applied. This is crucial for calculating integrals that may initially seem difficult or impossible.

Integral Calculus

Integral calculus is the process of finding the integral or antiderivative of functions. It includes a variety of techniques, such as the substitution method, which we used in this exercise.

• An integral essentially calculates the area under a curve. In mathematical terms, it is the reverse of differentiation.
• Integrals can be indefinite, as in this exercise, which includes a constant \( C \) to account for any constants lost during differentiation.
• Using the power rule for integration helps find antiderivatives for functions of form \( u^n \).

In this example, the integral of \( u^{-2} \) was evaluated using the power rule: the antiderivative is \( -u^{-1} \). This results in an expression that's easier to interpret or to be transformed back into the original variable.

Logarithmic Integration

Logarithmic integration comes into play when the integrand involves logarithmic expressions, such as \( \log_{8} x \) in this exercise. We see:

• Substitutions often use logarithmic forms to transform and simplify. Here, \( \log_{8} x \) becomes \( u \), which streamlines the integration process.
• Understanding properties such as the change of base formula \( \log_{8} x = \frac{\ln x}{\ln 8} \) is essential for simplifying and interpreting integral results.

After substituting back into terms of \( x \), we utilized this change of base formula. This reduced the expression to a simpler form. It demonstrated the elegance of logarithmic integration, making seemingly complex problems more approachable.