Evaluate the integrals in Exercises 25–30 by using a substitution prior to integration by parts.
∫ z(ln z)² dz

Verified step by step guidance

1. Identify the integral to solve: \(\int z (\ln z)^2 \, dz\).
2. Choose a substitution to simplify the integral. Let \(u = (\ln z)^2\). Then, find \(du\) in terms of \(dz\): since \(u = (\ln z)^2\), we have \(du = 2 \ln z \cdot \frac{1}{z} dz = \frac{2 \ln z}{z} dz\).
3. Rewrite the integral in terms of \(u\) and \(z\). Notice that the integral contains \(z (\ln z)^2 dz\), and from the substitution, express \(dz\) or parts of the integral accordingly to prepare for integration by parts.
4. Set up integration by parts. Choose \(U\) and \(dV\) such that the integral becomes easier to solve. For example, let \(U = (\ln z)^2\) and \(dV = z \, dz\), then compute \(dU\) and \(V\).
5. Apply the integration by parts formula: \(\int U \, dV = UV - \int V \, dU\). Substitute the expressions for \(U\), \(V\), and \(dU\) and simplify the resulting integral to complete the solution.

Key Concepts

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves choosing a substitution that transforms the integral into a more manageable form, often by setting a part of the integrand as a new variable. This technique is especially useful when the integral contains composite functions.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It is used to integrate products of functions by selecting parts of the integrand as 'u' and 'dv', then applying the formula ∫u dv = uv - ∫v du. This method is effective when the integrand is a product of algebraic and logarithmic or exponential functions.

Logarithmic Functions in Integration

Logarithmic functions, such as (ln z)², often require special techniques in integration due to their complexity. Recognizing how to handle powers of logarithms, often through substitution or integration by parts, is essential. Understanding their derivatives and integrals helps in simplifying and solving integrals involving logarithmic expressions.