Question

In the following exercises, verify by differentiation that ?lnxdx=x(lnx?1)+C, then use appropriate changes of variables to compute the integral. $$ \int x^{2} \ln ^{2} x d x $$

Short answer

The integral evaluates to \( x^3 \left( \frac{(\ln x)^2}{3} - \frac{2\ln x}{9} + \frac{2}{27} \right) + C \).

Step-by-step solution

Verify by Differentiation

To verify the antiderivative \( \int \ln x \, dx = x(\ln x - 1) + C \), differentiate \( x(\ln x - 1) \).Use the product rule: \( \frac{d}{dx}[x(\ln x - 1)] = \frac{d}{dx}[x] \cdot (\ln x - 1) + x \cdot \frac{d}{dx}[\ln x - 1] \).This simplifies to \( 1(\ln x - 1) + x \cdot (\frac{1}{x}) = \ln x - 1 + 1 = \ln x \).Thus, \( \int \ln x \, dx = x(\ln x - 1) + C \) is verified.

Set Up the Integral with a Change of Variables

To compute \( \int x^2 \ln^2 x \, dx \), use substitution.Let \( u = \ln x \). Then \( \frac{du}{dx} = \frac{1}{x} \Rightarrow du = \frac{1}{x} dx \).Therefore, \( dx = x \, du \), and since \( x = e^u \), we re-write our integral as \( \int (e^u)^2 u^2 (e^u \, du) = \int e^{3u} u^2 \, du \).

Integration by Parts

Use integration by parts formula \( \int u \, dv = uv - \int v \, du \) on \( \int e^{3u} u^2 \, du \).Let \( v = e^{3u} \) and \( w = u^2 \), then \( dv = 3e^{3u} \, du \) and \( dw = 2u \, du \).Perform integration by parts:1. \( \int u^2 \cdot e^{3u} \, du = u^2 \cdot \frac{1}{3}e^{3u} \bigg| - \int \frac{2u}{3}e^{3u} \, du \).2. Further simplify using a second iteration of integration by parts for the remaining integral.

Execute the Second Iteration

For \( \int u e^{3u} \, du \), again apply integration by parts using \( v = e^{3u} \) and \( w = u \), thus \( dv = 3e^{3u} \, du \) and \( dw = du \).The integration process is:1. \( ue^{3u} \cdot \frac{1}{3} \bigg| - \int e^{3u} \cdot \frac{1}{3} \, du \).2. Integrating yields: \( \frac{1}{3}ue^{3u} - \frac{1}{9}e^{3u} \).

Evaluate and Simplify

Combine results from both integration steps:1. \( \frac{u^2}{3} e^{3u} - \int \frac{2}{3} (\frac{1}{3}ue^{3u} - \frac{1}{9}e^{3u}) \, du \).2. Incorporate into the original integral solving process and replace the substitution back with \( u = \ln x \).3. Simplify the expression to evaluate the definite integral.

Key concepts

Differentiation Verification

Differentiation is a crucial tool in calculus that helps us verify whether a given function is indeed the antiderivative of another. In our problem, we started by verifying that the antiderivative of the integral \( \int \ln x \, dx \) is \( x(\ln x - 1) + C \). This verification is achieved by differentiating the result \( x(\ln x - 1) \). By using the product rule, which states that \( \frac{d}{dx} [uv] = u'v + uv' \), we differentiated \( x(\ln x - 1) \) and saw that it returns \( \ln x \), confirming our antiderivative is correct.

• The product rule helps break down complex derivatives into manageable parts.
• Once verified, this helps in ensuring the antiderivative you derived is accurate.

Understanding this process strengthens our grasp on how integration and differentiation are inverse processes.Keep practicing different examples to get comfortable with differentiation verification.

Change of Variables

The technique of change of variables is a powerful method in integration, particularly useful when dealing with complex integrals. This technique involves substituting a part of the integral with a new variable, simplifying the integral into a more manageable form. In our problem, we tackled the integral \( \int x^2 \ln^2 x \, dx \) by letting \( u = \ln x \).This substitution turned our integral into \( \int e^{3u} u^2 \, du \), which is simpler to integrate. We used:

• \( \frac{du}{dx} = \frac{1}{x} \to du = \frac{1}{x} \, dx \)
• Substituted \( x = e^u \) to match the new integral form.

Such substitutions make it easier to handle integrals that are otherwise complicated or resistant to standard techniques.Practicing this method enhances one's ability to see problems from different perspectives and simplifies solving integrals.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. This method is especially useful when integrating the product of two functions, such as \( \int e^{3u} u^2 \, du \) encountered in our exercise. Here, we used the formula \( \int u \, dv = uv - \int v \, du \).To apply integration by parts:

• Select which functions to assign \( u \) and \( dv \). In our solution, we took \( v = e^{3u} \) and \( w = u^2 \).
• Set the derivative parts: \( dv = 3e^{3u} \, du \) and \( dw = 2u \, du \).
• Perform the integration step-by-step, integrating piece by piece.

This method often requires multiple iterations, as seen in our subsequent integration by parts steps for \( \int u e^{3u} \, du \). Integration by parts, when practiced, becomes an invaluable tool for dealing with otherwise challenging integrals.Understanding its applications across various problems will empower you to tackle diverse calculus challenges with confidence.