Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 49


$$\frac{\sqrt{x^{2}+4}}{16 x}-\frac{\left(x^{2}+4\right) \sqrt{x^{2}+4}}{48 x^{3}}+C$$

Work Step by Step

Given $$\int \frac{d x}{x^{4} \sqrt{x^{2}+4}}$$ Let $$x=2\tan u\ \ \ \ \ \ \ \ dx=2\sec^2 udu $$ Then \begin{align*} \int \frac{d x}{x^{4} \sqrt{x^{2}+4}}&=\int \frac{2\sec^2 udu }{16\tan ^{4} \sqrt{4\tan^{2}u+4}}\\ &=\int \frac{\sec^2 udu }{8\tan ^{4} \sqrt{4\sec^{2}u}}\\ &=\frac{1}{16}\int \frac{\sec udu }{\tan ^{4} }\\ &=\frac{1}{16}\int \cot^3 u \csc udu\\ &=\frac{1}{16}\int (1-\csc^2 u)\cot u \csc udu\\ &=\frac{1}{16} \left( -\csc u+\frac{1}{3}\csc^3u\right)+C\\ &= \frac{\sqrt{x^{2}+4}}{16 x}-\frac{\left(x^{2}+4\right) \sqrt{x^{2}+4}}{48 x^{3}}+C \end{align*}
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