University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 8


$$\int \frac{1}{x^2\sqrt{4x-9}}dx=\frac{\sqrt{4x-9}}{9x}+\frac{4}{27}\tan^{-1}\sqrt{\frac{4x-9}{9}}+C$$

Work Step by Step

$$A=\int \frac{1}{x^2\sqrt{4x-9}}dx$$ Using Formula 31, which states that $$\int \frac{dx}{x^2\sqrt{ax+b}}=-\frac{\sqrt{ax+b}}{bx}-\frac{a}{2b}\int\frac{dx}{x\sqrt{ax+b}}+C$$ with $a=4$ and $b=-9$ here, we have $$A=-\frac{\sqrt{4x-9}}{-9x}-\frac{4}{-18}\int\frac{dx}{x\sqrt{4x-9}}+C$$ $$A=\frac{\sqrt{4x-9}}{9x}+\frac{2}{9}\int\frac{dx}{x\sqrt{4x-9}}+C$$ For the remaining integral, we apply Formula 29b, which states that $$\int\frac{dx}{x\sqrt{ax-b}}=\frac{2}{\sqrt b}\tan^{-1}\sqrt{\frac{ax-b}{b}}+C$$ with $a=4$ and $b=9$: $$A=\frac{\sqrt{4x-9}}{9x}+\frac{2}{9}\Big(\frac{2}{3}\tan^{-1}\sqrt{\frac{4x-9}{9}}\Big)+C$$ $$A=\frac{\sqrt{4x-9}}{9x}+\frac{4}{27}\tan^{-1}\sqrt{\frac{4x-9}{9}}+C$$
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