Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.1 - Using Basic Integration Formulas - Exercises 8.1 - Page 448: 2


$$\int \frac{x^{2}}{x^{2}+1}dx=x-arctan\,x+C$$

Work Step by Step

$$\int \frac{x^{2}}{x^{2}+1}dx=\int \frac{x^{2}+1-1}{x^{2}+1}dx$$ $$=\int (1-\frac{1}{x^{2}+1})dx$$ $$=\int 1dx-\int \frac{1}{x^{2}+1}dx$$ $$=x-arctan\,x+C$$
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