Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 91


$- \sqrt{4-x^2} +5\sin^{-1}(x/2)+c.$

Work Step by Step

We have $$\int \frac{(x+5)dx}{\sqrt{4-x^2}}=\int \frac{xdx}{\sqrt{4-x^2}}+\int \frac{5dx}{\sqrt{4-x^2}}\\ =-\frac{1}{2}\int \frac{-2xdx}{\sqrt{4-x^2}}+5\int \frac{d(x/2)}{\sqrt{1-(x/2)^2}}\\ =-\frac{1}{2} \frac{\sqrt{4-x^2}}{1/2}+5\sin^{-1}(x/2)+c\\ =- \sqrt{4-x^2} +5\sin^{-1}(x/2)+c.$$
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