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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.5.41

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

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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.

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Key Concepts

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