Skip to main content
Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.5.80
Capitolo 5, Problema 5.5.80

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

Guida verificata passo dopo passo
1
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.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.
Video consigliato:
07:33
Euler's Method

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.
Video consigliato:
Percorso guidato
06:18
Integration by Parts for Definite Integrals

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.
Video consigliato:
07:00
Taylor Polynomials
Pratica correlata
Domanda del libro di testo

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

69
views
Domanda del libro di testo

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

                                                                                                                                                                              

 ∫₀¹ 2e²ˣ d𝓍

74
views
Domanda del libro di testo

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

157
views
Domanda del libro di testo

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
Domanda del libro di testo

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

                                                                                                                                                                              

 ∫₁³ ( 2ˣ / 2ˣ + 4 ) d𝓍

42
views
Domanda del libro di testo

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

70
views