Problem 98 - Calculus and its applications

Question

Evaluate. $$ \int \frac{(\ln x)^{99}}{x} d x, x>0 $$

Step-by-Step Solution

Answer

\( \frac{(\ln x)^{100}}{100} + C \)

Step 1: Recognize the Integral Type

This integral is a case for substitution due to the presence of a logarithmic function raised to a power in the numerator and its derivative component, \( \frac{1}{x} \), in the denominator.

Step 2: Choose the Appropriate Substitution

Let \( u = \ln x \). Thus, the derivative \( \frac{du}{dx} = \frac{1}{x} \), which implies \( du = \frac{1}{x} dx \). Substitute these into the integral.

Step 3: Substitute and Simplify the Integral

After substitution, the integral becomes \( \int u^{99} \, du \). This simplification is achieved using the substitution \( u = \ln x \) and \( du = \frac{1}{x} dx \).

Step 4: Integrate Using the Power Rule

Apply the power rule for integration to \( \int u^{99} \, du = \frac{u^{100}}{100} + C \), where \( C \) is the constant of integration.

Step 5: Substitute Back to the Original Variable

Re-substitute \( u = \ln x \) back into the expression. Therefore, the integral becomes \( \frac{(\ln x)^{100}}{100} + C \).

Key Concepts

Understanding Substitution in Integral Calculus

When faced with complicated integrals like \( \int \frac{(\ln x)^{99}}{x} \, dx \), substitution is a powerful technique to simplify the problem. The idea is to change variables to transform the integral into an easier form. Here's how:

• Identify a function within the integral whose derivative also appears in the integrand. In this case, \( \ln x \) is our candidate because its derivative \( \frac{1}{x} \) is part of the integrand.
• Substitute the complex part, \( \ln x \), with a simpler term, \( u \). This gives \( u = \ln x \).
• Differentiate \( u \) to find \( du \), resulting in \( du = \frac{1}{x} dx \). Notice that \( \frac{1}{x} dx \) perfectly matches a part of the integral, making substitution seamless.

This substitution transforms our original integral into \( \int u^{99} \, du \), a much simpler expression that we can handle with basic integration rules.

Applying the Power Rule to Integrate

Once substitution has narrowed down the integral to \( \int u^{99} \, du \), we can use the power rule of integration. This rule is straightforward:

• Formula: For any function \( u^n \), where \( n eq -1 \), the integral is \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \).
• In our problem, \( n = 99 \), so applying the power rule gives us \( \int u^{99} \, du = \frac{u^{100}}{100} + C \).
• Constant of Integration: The term \( C \) is crucial. When integrating, always include the integration constant to account for the family of possible solutions.

By integrating \( u^{99} \), we have moved closer to our solution. The power rule makes this step of the problem-solving process much less daunting.

Exploring Logarithmic Functions in Calculus

Logarithmic functions, like \( \ln x \), play a significant role in calculus, particularly in integration and differentiation. Let's consider why they are special:

• Derivative: The derivative of \( \ln x \) is \( \frac{1}{x} \), a simple expression that frequently appears in integrals that involve substitution.
• Integration: Integration involving \( \ln x \) often requires creative approaches such as substitution to simplify the process.
• Properties: Logarithmic functions possess unique properties, like \( \ln(ab) = \ln a + \ln b \), which can be helpful when simplifying expressions.

In our integral problem, \( \ln x \) served as a brilliant choice for substitution due to its straightforward derivative and the presence of the derivative part \( \frac{1}{x} \) in the integrand. Logarithmic functions continue to be an essential tool in both theoretical and applied calculus problems.