## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Practice Exercises - Page 474: 24

#### Answer

$x^2-3x+\dfrac{2}{3} \ln |x+4|+\dfrac{1}{3} \ln |x-2|+C$

#### Work Step by Step

Consider the integral $\int \dfrac{2x^3+x^2-21x+24}{x^2+2x-8}dx$ The given integral can be re-written as: $\int \dfrac{2x^3+x^2-21x+24}{x^2+2x-8}dx=\int (2x-3)+ \dfrac{x}{x^2+2x-8}dx$ This implies that $\int [(2x-3)+ \dfrac{x}{x^2+2x-8}]dx=\int (2x-3) dx+(\dfrac{1}{3})\int \dfrac{dx}{x-2}+(\dfrac{2}{3}) \int \dfrac{dx}{x+4}$ We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$ or, $=x^2-3x+\dfrac{2}{3} \ln |x+4|+\dfrac{1}{3} \ln |x-2|+C$

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