Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.PE.61
Chapter 7, Problem 7.PE.61

Evaluate the integrals in Exercises 31–78.
61. ∫(from 1 to 3)(ln(v+1))²/(v+1) dv

Verified step by step guidance
1
Recognize that the integral is of the form \( \int_{1}^{3} \frac{(\ln(v+1))^2}{v+1} \, dv \). To simplify, use the substitution \( u = v + 1 \).
Compute the differential: since \( u = v + 1 \), then \( du = dv \). Also, change the limits of integration accordingly: when \( v = 1 \), \( u = 2 \); when \( v = 3 \), \( u = 4 \).
Rewrite the integral in terms of \( u \): \( \int_{2}^{4} \frac{(\ln u)^2}{u} \, du \).
Recognize that this integral can be approached by using the substitution \( t = \ln u \), which implies \( dt = \frac{1}{u} du \). This transforms the integral into \( \int_{t=\ln 2}^{t=\ln 4} t^2 \, dt \).
Integrate \( t^2 \) with respect to \( t \) over the new limits, then substitute back if needed to express the answer in terms of \( v \).

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 substituting a part of the integrand with a new variable to transform the integral into a more manageable form. For example, setting u = v + 1 can simplify the given integral by reducing the complexity of the integrand.
Recommended video:
04:27
Substitution With an Extra Variable

Properties of Logarithmic Functions

Understanding logarithmic functions, especially natural logarithms (ln), is essential. The function ln(x) is the inverse of the exponential function e^x, and its properties, such as the chain rule in differentiation and integration, help in manipulating integrals involving ln expressions.
Recommended video:
06:21
Properties of Functions

Definite Integrals and Limits of Integration

Definite integrals calculate the net area under a curve between two points. Evaluating a definite integral requires applying the Fundamental Theorem of Calculus, which involves finding an antiderivative and then computing the difference of its values at the upper and lower limits.
Recommended video:
05:43
Definition of the Definite Integral
Related Practice
Textbook Question

Evaluate the integrals in Exercises 31–78.

31. ∫e^x sin(e^x)dx

Textbook Question

Use l’Hôpital’s Rule to find the limits in Exercises 85–108.

93. lim(x→0) (csc(x) - cot(x))

Textbook Question

110. Does f grow faster, slower, or at the same rate as g as x→∞? Give reasons for your answers.

a. f(x) = 3^(-x), g(x) = 2^(-x)

Textbook Question

Evaluate the integrals in Exercises 31–78.

35. ∫sec²x e^(tan x)dx

Textbook Question

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.

9. y = 8^(-t)

Textbook Question

Evaluate the integrals in Exercises 31–78.

58. ∫(from 0 to ln9)e^θ(e^θ-1)^(1/2) dθ