## Calculus 10th Edition

Published by Brooks Cole

# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 19

#### Answer

$\frac{1}{2}(ln|(x^2+2)|)+\frac{1}{3}x^3-2x+C$

#### Work Step by Step

$\int \frac{x^4+x-4}{x^2+2} dx$ //divide the numerator with the denominator $=\int(\frac{x}{x^2+2}+x^2-2)dx$ $=\int\frac{x}{x^2+2}dx+ \int x^2 dx - \int 2dx$ let $x^2+2=u$ $2x$ $dx$$=du$ $=\int\frac{x}{x^2+2}dx$ $=\frac{1}{2} \int \frac{1}{u}du$ $=\frac{1}{2}(ln|u|)$ $=\frac{1}{2}(ln|(x^2+2)|)$ $=\int\frac{x}{x^2+2}dx+ \int x^2 dx - \int 2dx$ $=\frac{1}{2}(ln|(x^2+2)|)+\frac{1}{3}x^3-2x+C$

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