## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 17

#### Answer

$$\frac{1}{4} \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+C$$

#### Work Step by Step

Given $$\int \frac{d x}{x \sqrt{x^{2}+16}}$$ Let $$x= 4\tan \theta ,\ \ \ dx= 4\sec^2\theta d\theta$$ Then \begin{aligned} \int \frac{d x}{x \sqrt{x^{2}+16}} &=\int \frac{4 \sec ^{2} \theta d \theta}{4 \tan \theta \sqrt{(4 \tan \theta)^{2}+16}} \\ &=\int \frac{A \sec ^{2} \theta d \theta}{A \tan \theta \sqrt{16 \tan ^{2} \theta+16}} \\ &=\int \frac{\sec ^{2} \theta d \theta}{4 \tan \theta \sqrt{\sec ^{2} \theta}} \\ &=\frac{1}{4} \int \frac{\sec ^{2} \theta d \theta}{\tan \theta \sec \theta} \\ &=\frac{1}{4} \int \frac{\sec \theta d \theta}{\tan \theta} \\ &=\frac{1}{4} \int \csc \theta d \theta \\ &= \frac{1}{4}\ln | \csc \theta -\cos\theta |\\ &= \frac{1}{4} \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+C \end{aligned}

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