Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 45

Answer

$$\int\frac{1+x}{1+x^2}dx=\tan^{-1}x+\frac{1}{2}\ln|1+x^2|+C$$

Work Step by Step

$$A=\int\frac{1+x}{1+x^2}dx$$ Sometimes, the numerator of the fraction inside the integral might contain several elements like this one. You might try to break down the elements into several integrals and see if it would be easier to transform to familiar forms or not. $$A=\int [\frac{1}{1+x^2}+\frac{x}{1+x^2}]dx$$ $$A=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx$$ $$A=B+C$$ *Consider $B$: $$B=\int\frac{1}{1+x^2}dx$$ $$B=\tan^{-1}x+C$$ *Consider $C$: $$C=\int\frac{x}{1+x^2}dx$$ Let $u=1+x^2$ Then we have $du=2xdx$. That means $xdx=\frac{1}{2}du$ Substitute into $C$: $$C=\frac{1}{2}\int\frac{1}{u}du$$ $$C=\frac{1}{2}\ln|u|+C$$ $$C=\frac{1}{2}\ln|1+x^2|+C$$ Overall, $$A=\tan^{-1}x+\frac{1}{2}\ln|1+x^2|+C$$
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