Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.5.80

Chapter 5, Problem 5.5.80

Variations on the substitution method Evaluate the following integrals.

∫ y²/(y + 1)⁴ dy

Verified step by step guidance

Step 1: Identify the substitution method. Let u = y + 1, which simplifies the denominator. Then, du = dy.

Step 2: Rewrite y in terms of u. Since u = y + 1, we have y = u - 1.

Step 3: Substitute into the integral. Replace y² with (u - 1)² and (y + 1)⁴ with u⁴, and dy with du. The integral becomes ∫ (u - 1)² / u⁴ du.

Step 4: Expand (u - 1)² to simplify the numerator. This gives (u - 1)² = u² - 2u + 1. Substitute this into the integral, resulting in ∫ (u² - 2u + 1) / u⁴ du.

Step 5: Break the integral into separate terms. Divide each term in the numerator by u⁴: ∫ (u²/u⁴ - 2u/u⁴ + 1/u⁴) du, which simplifies to ∫ (1/u² - 2/u³ + 1/u⁴) du. Now, integrate each term individually.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Substitution Method

The substitution method is a technique used in integration to simplify the process by changing the variable of integration. By substituting a part of the integrand with a new variable, the integral can often be transformed into a more manageable form. This method is particularly useful when dealing with composite functions or when the integrand contains a function and its derivative.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation, used to integrate products of functions. It states that the integral of a product of two functions can be expressed as the product of one function and the integral of the other, minus the integral of the derivative of the first function times the second. This method is helpful when the integrand is a product that can be simplified through differentiation.

Polynomial Long Division

Polynomial long division is a method used to divide a polynomial by another polynomial of lower degree. This technique is particularly useful when the integrand is a rational function, allowing the integrand to be expressed as a sum of a polynomial and a proper fraction. By simplifying the integrand in this way, it becomes easier to integrate each term separately, facilitating the overall integration process.

Recommended video:

07:00

Taylor Polynomials

Related Practice

Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.

∫ sin 𝓍 sec⁸ 𝓍 d𝓍

views

Textbook Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.

∫₀¹ 2e²ˣ d𝓍

views

Textbook Question

Why can the constant of integration be omitted from the antiderivative when evaluating a definite integral?

157

views

Textbook Question

Average distance on a parabola What is the average distance between the parabola y = 30𝓍 (20 ― 𝓍 ) and the 𝓍-axis on the interval [0, 20] ?

123

views

Textbook Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.

∫₁³ ( 2ˣ / 2ˣ + 4 ) d𝓍

views

Textbook Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.

∫π/₄^π/² (cos 𝓍) / (sin² 𝓍) d𝓍

views

Variations on the substitution method Evaluate the following integrals. ∫ y²/(y + 1)⁴ dy

Verified video answer for a similar problem:

Key Concepts

Substitution Method

Integration by Parts

Polynomial Long Division

Variations on the substitution method Evaluate the following integrals.

∫ y²/(y + 1)⁴ dy