Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.1 - Using Basic Integration Formulas - Exercises 8.1 - Page 448: 28


$$\int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}} =\sec ^{-1}(x-2)+c $$

Work Step by Step

Given $$\int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}}$$ So, we get \begin{aligned} I&= \int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}}\\ &= \int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+4-1}}\\ &=\int \frac{d x}{(x-2) \sqrt{(x-2)^{2}-1}} \\ \end{aligned} Let $$\begin{array}{l}{x-2=\sec \theta} \\ {\Rightarrow d x=\sec \theta \tan \theta d \theta}\end{array}$$ So, we get \begin{aligned} I&= \int \frac{\sec \theta \tan \theta \ \ d \theta}{\sec \theta \sqrt{\sec ^{2} \theta-1}}\\ &=\int \frac{\sec \theta \tan \theta \ \ d \theta}{\sec \theta \tan \theta} \\ & =\int d \theta\\ &=\theta+c\\ &=\sec ^{-1}(x-2)+c \end{aligned}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.