Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 27


\begin{aligned} \int\sec ^{-1} x d x= x \sec ^{-1} x-\ln |x+\sqrt{x^{2}-1}|+C \end{aligned}

Work Step by Step

Given $$ \int \sec ^{-1} x d x $$ Use integration by parts: $$ u=\sec ^{-1} x \Rightarrow du= \frac{1}{x\sqrt {x^2-1}}dx $$ $$ dv= dx \Rightarrow v= x $$ So, we get \begin{aligned} I&=\int\sec ^{-1} x d x\\ &=uv - \int vdu\\ &=x\sec ^{-1} x-\int \frac{x}{x\sqrt {x^2-1}}dx\\ &=x\sec ^{-1} x-\int \frac{1}{\sqrt {x^2-1}}dx\\ &=x\sec ^{-1} x-\ln |x+\sqrt{x^{2}-1}|+C \end{aligned} Check: $$ \begin{aligned} \frac{d}{d x}[\ln |x+\sqrt{x^{2}-1}|] &=\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(1+\frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x\right) \\ &=\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(1+\frac{x}{\sqrt{x^{2}-1}}\right) \\ &=\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(\frac{\sqrt{x^{2}-1}+x}{\sqrt{x^{2}-1}}\right) \\ &=\frac{1}{\sqrt{x^{2}-1}} \end{aligned} $$
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