Skip to main content
Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.5.41

23-64. Integration Evaluate the following integrals.
41. ∫₋₁¹ x/(x + 3)² dx

Guida verificata passo dopo passo
1
First, observe the integral \( \int_{-1}^{1} \frac{x}{(x + 3)^2} \, dx \). Notice that the integrand is a rational function where the denominator is \( (x+3)^2 \).
To simplify the integral, consider using substitution. Let \( u = x + 3 \). Then, \( du = dx \) and \( x = u - 3 \). Also, change the limits of integration accordingly: when \( x = -1 \), \( u = 2 \), and when \( x = 1 \), \( u = 4 \).
Rewrite the integral in terms of \( u \): \( \int_{2}^{4} \frac{u - 3}{u^2} \, du \). This can be separated into two simpler integrals: \( \int_{2}^{4} \frac{u}{u^2} \, du - 3 \int_{2}^{4} \frac{1}{u^2} \, du \).
Simplify the integrands: \( \frac{u}{u^2} = \frac{1}{u} \) and \( \frac{1}{u^2} = u^{-2} \). So the integral becomes \( \int_{2}^{4} \frac{1}{u} \, du - 3 \int_{2}^{4} u^{-2} \, du \).
Now, integrate each term separately: \( \int \frac{1}{u} \, du = \ln|u| \) and \( \int u^{-2} \, du = -u^{-1} \). After integrating, apply the limits \( 2 \) to \( 4 \) to find the definite integral.

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:
8m

Concetti chiave

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

Definite Integration

Definite integration calculates the net area under a curve between two limits. It involves evaluating the integral of a function from a lower to an upper bound, resulting in a numerical value representing accumulated quantity or area.
Video consigliato:
Percorso guidato
05:43
Definition of the Definite Integral

Integration by Substitution

Integration by substitution simplifies integrals by changing variables to transform the integral into a more manageable form. It is useful when the integrand contains a composite function, allowing easier integration after substitution.
Video consigliato:
04:27
Substitution With an Extra Variable

Handling Rational Functions

Rational functions are ratios of polynomials, often requiring algebraic manipulation or substitution for integration. Recognizing patterns like derivatives of denominators in the numerator helps in applying substitution or partial fraction techniques.
Video consigliato:
6:04
Intro to Rational Functions