Problem 30 - Thomas Calculus

Question

Evaluate the integrals by using a substitution prior to integration by parts. \(\int z(\ln z)^{2} d z\)

Step-by-Step Solution

Answer

\( \int z(\ln z)^2 \, dz = z(\ln z)^2 - 2z(\ln z) + 2z + C \).

Step 1: Identify Suitable Substitution

Let's choose a substitution that simplifies the integrand. Notice that we have a term \((\ln z)^2\) which suggests the natural logarithm might be a good candidate for substitution. Let's set \( u = \ln z \). This gives us \( du = \frac{1}{z} dz \) or \( dz = z\, du \).

Step 2: Substitute and Simplify the Integral

Perform the substitution using \( u = \ln z \) and \( dz = z \cdot du \), which modifies the integral as follows:\[\int z (\ln z)^2 \, dz = \int u^2 \, z \, du.\]Substituting \( z = e^u \), the integral becomes:\[\int u^2 e^u \, du.\]

Step 3: Apply Integration by Parts

We will use integration by parts, which states \( \int u \, dv = uv - \int v \, du \). Let \( u = u^2 \) and \( dv = e^u \, du \). Then \( du = 2u \, du \) and \( v = e^u \).

Step 4: Differentiate and Integrate parts

Substitute back into the integration by parts formula:\[\int u^2 e^u \, du = u^2 e^u - \int 2u e^u \, du.\]This requires a second application of integration by parts on \( \int 2u e^u \, du \). Use \( u = 2u \) and \( dv = e^u \, du \), producing \( du = 2 \, du \) and \( v = e^u \).

Step 5: Evaluate the Remaining Integral

Apply integration by parts again to \( \int 2u e^u \, du \):\[\int 2u e^u \, du = 2u e^u - \int 2e^u \, du = 2u e^u - 2e^u.\]Substitute this back:\[\int u^2 e^u \, du = u^2 e^u - (2u e^u - 2e^u) = u^2 e^u - 2u e^u + 2e^u.\]

Step 6: Back Substitute in Terms of Original Variable

Revert the substitution with \( u = \ln z \):\[(u^2 e^u - 2u e^u + 2e^u) = ((\ln z)^2 z - 2(\ln z) z + 2z).\]Thus, the integral in terms of \( z \) is:\[\int z(\ln z)^2 \, dz = z(\ln z)^2 - 2z(\ln z) + 2z + C,\]where \( C \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in solving integrals, particularly when dealing with more complicated functions. The idea is to simplify the integral by changing the variable of integration to something easier to work with. For this given problem, we have the integral \( \int z(\ln z)^2 \, dz \). Here, we decide to apply the substitution method with the hope of simplifying it before tackling integration by parts.

A good choice for substitution is selecting parts of the integrand that are more complex. In this case, noticing \((\ln z)^2\) indicates that \( \ln z \) might be a good candidate. Therefore, we set \( u = \ln z \). This choice then requires calculating the corresponding differential: \( du = \frac{1}{z} dz \), which rearranges to \( dz = z \, du \).

By substituting these into the integral, you get \( \int u^2 z \, du \). However, we know that \( z = e^u \) from our substitution, which transforms the integral into \( \int u^2 e^u \, du \). This new integral is more manageable and sets us up perfectly to apply the method of integration by parts.

Natural Logarithm Integration

Integrating functions that involve the natural logarithm, such as \( \ln z \), requires special techniques because the logarithm itself and its powers can complicate the integration process. In the context of this exercise, we had \( \int z(\ln z)^2 \, dz \), which indeed involves the natural logarithm squared.

The natural logarithm often appears in problems that can be simplified using substitution. However, when faced with powers or products of \( \ln z \), like \( (\ln z)^2 \), additional strategies such as integration by parts become necessary. These techniques help to break down the complexity by reducing the degree of polynomials or isolating the logarithmic component.

In many cases, handling \( (\ln z)^{n} \) in integration requires several iterations of integration techniques, which might include repeated use of integration by parts or clever substitutions that redefine the original variable. Understanding how the logarithm interacts with these techniques is crucial for efficiently solving related integrals.

Integration Techniques

Various integration techniques, such as substitution and integration by parts, are essential tools for tackling problems where straightforward integration isn't possible. Our initial task was to evaluate \( \int z(\ln z)^2 \, dz \), which isn't straightforward. So, to manage it, we use combinations of these powerful techniques.

After using the substitution method to rewrite the integral in terms of \( u \) and \( e^u \), we are left with \( \int u^2 e^u \, du \). Integration by parts is now appropriate. This technique is based on the product rule for differentiation and is useful when you have products of functions, like polynomials multiplied by exponentials or logarithms.

Integration by parts formula, \( \int u \, dv = uv - \int v \, du \), allows us to peel away layers of the integrand, reducing the degree of complexity with each application. Multiple applications may be necessary, as seen in this problem, where it required a second round of integration by parts to fully evaluate the integral. Mastering these techniques is key in handling intricate integrals efficiently.