Problem 24 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int(\ln x)^{2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( x((\ln x)^2 - 2 \ln x + 2) + C \) where \( C \) is the integration constant.

Step 1: Choose Integration Technique

To solve \( \int (\ln x)^2 \, dx \), we should start by selecting an integration technique. Given the complexity of the integrand, using integration by parts is appropriate. Integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 2: Assign Functions for Integration by Parts

Set \( u = (\ln x)^2 \) and \( dv = dx \). This implies that \( du = 2 \ln x \cdot \frac{1}{x} \, dx \) and \( v = x \).

Step 3: Apply Integration by Parts Formula

Using the integration by parts formula, we have:\[\int (\ln x)^2 \, dx = x (\ln x)^2 - \int x \cdot 2 \ln x \cdot \frac{1}{x} \, dx\]This simplifies to:\[x (\ln x)^2 - 2\int \ln x \, dx\]

Step 4: Integrate the Remaining Integral

To solve \( \int \ln x \, dx \), apply integration by parts again. Let \( u = \ln x \) and \( dv = dx \). Then, \( du = \frac{1}{x} \, dx \) and \( v = x \). So:\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x\]

Step 5: Substitute Back into Original Equation

Substitute the result from Step 4 back into the equation from Step 3:\[x (\ln x)^2 - 2 (x \ln x - x) = x (\ln x)^2 - 2x \ln x + 2x\]

Step 6: Simplify the Result

Simplify the expression:\[x (\ln x)^2 - 2x \ln x + 2x = x((\ln x)^2 - 2 \ln x + 2)\]

Step 7: Differentiate to Check the Solution

Differentiate \( x((\ln x)^2 - 2 \ln x + 2) \) with respect to \( x \) to verify:\[d\left(x((\ln x)^2 - 2 \ln x + 2)\right)/dx = (\ln x)^2 - 2 \ln x + 2 + 2 \ln x - 2 = (\ln x)^2 \]The differentiation returns the original integrand, confirming the solution is correct.

Key Concepts

Integration Technique

Integration by parts is a powerful tool when dealing with integrals where the standard methods of integration don't easily apply. This technique helps us split a complicated integrand into more manageable parts. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

The goal is to choose appropriate functions for \( u \) and \( dv \) so that the derivative \( du \) and the integrable function \( v \) simplify the original integral. In our exercise, for the integral \( \int (\ln x)^2 \, dx \), the choice was to let \( u = (\ln x)^2 \) and \( dv = dx \). This choice allows us to break down the complexity of the logarithmic squared function and work through the integration more easily. The challenge lies in strategically selecting \( u \) and \( dv \) to ensure that \( du \) and \( \int v \, du \) remain integrable. This technique is usually applied in cases where other simpler techniques are ineffective, as it deconstructs the problem into smaller, familiar parts.

Differentiation

Differentiation is the mathematical process of finding a derivative, which measures how a function changes as its input changes. This is essential not only in checking the correctness of a solution—like we did in the exercise by differentiating the final expression—but also in understanding how parts of a function behave. When using integration by parts, we need to calculate the derivative \( du \) of the chosen \( u \).

• For \( u = (\ln x)^2 \), the derivative \( du \) involves applying the chain rule: \( du = 2 \ln x \cdot \frac{1}{x} \, dx \).

Differentiation allows us to verify our integration by parts result by ensuring that when we differentiate the final expression, we retrieve the original integrand: \((\ln x)^2\). This step acts as a perfect check for our work, confirming the accuracy of our solution.

Integrand

The integrand is the function we wish to integrate, and understanding its structure can often guide us in choosing the appropriate integration technique. In this exercise, the integrand \((\ln x)^2\) presents as a complex logarithmic expression raised to a power, making it a prime candidate for integration by parts. The integrand can be intimidating at first due to its form, but breaking it down into parts—such as separating the \( \ln x \) for use in integration by parts—unlocks a path to simplify the problem.

• Recognizing patterns, like the squared logarithmic form here, can determine the most efficient method, eliminating the trial and error of testing other methods.

This strategic approach minimizes unnecessary complexity and helps to manage what could seem like an overwhelmingly complicated integral at first.

Substitution Method

The substitution method is another fundamental integration technique, especially useful when the integral contains a function and its derivative. Though not directly used in the main solution here, substitution is worth understanding for its complementary nature to integration by parts. It often serves where integration by parts might not directly simplify the problem enough. The substitution method involves:

• Changing variables to simplify the integrand, turning it into a more standard or easier-to-integrate form.
• Choosing a substitution that conveniently transforms the limits and the differential \( dx \) into a new variable \( du \).

While this exercise primarily relied on integration by parts, knowing when to use substitution—typically when recalculating smaller integrations—is handy. Even within integration by parts applications, substitution may aid in intermediate steps. Its role in simplifying parts of an integrand can substantially reduce the time and effort required to reach a solution, making it a crucial technique to have in your toolkit.