Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.2.30
Chapter 8, Problem 8.2.30

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: \(\int z (\ln z)^2 \, dz\).
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 \(du = 2 \ln z \cdot \frac{1}{z} dz = \frac{2 \ln z}{z} dz\).
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.
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 \(V\).
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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
Recommended video:
04:27
Substitution With an Extra Variable

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.
Recommended video:
06:18
Integration by Parts for Definite Integrals

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.
Recommended video:
5:26
Graphs of Logarithmic Functions
Related Practice
Textbook Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 0 to ∞ of (dθ / (1 + e^θ))

1
views
Textbook Question

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.

∫₁^∞ (1 / x^(1/5)) dx

Textbook Question

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.

∫ (dθ / √(2θ - θ²))

Textbook Question

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.

∫ 3 sinh(x/2 + ln 5) dx

1
views
Textbook Question

Use integration by parts to obtain the formula ∫ √(1 - x²) dx = (1/2) x √(1 - x²) + (1/2) ∫ 1 / √(1 - x²) dx.

Textbook Question

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.

∫ (dt / t√(3 + t²)

1
views