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: 33


\begin{aligned} \int \tanh ^{-1}4 x \ dx =x\tanh ^{-1}4 x+\frac{1}{8} \ln|1-16x^2| +C\\ \end{aligned}

Work Step by Step

Given $$\int \tanh ^{-1}4 x \ dx $$ Use integration by parts: $$ u=\tanh ^{-1} 4x \Rightarrow du= \frac{4}{1-16x^2}dx $$ $$ dv= dx \Rightarrow v=x $$ So, we get \begin{aligned} I&=\int \tanh ^{-1}4 x \ dx\\ &=uv- \int vdu\\ &=x\tanh ^{-1}4 x- \int \frac{4x}{1-16x^2}d x\\ &=x\tanh ^{-1}4 x+\frac{1}{8} \int \frac{-32x}{1-16x^2} d x\\ &=x\tanh ^{-1}4 x+\frac{1}{8} \ln|1-16x^2| +C\\ \end{aligned}
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