Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 73


$$\int\frac{(\arctan x)^2}{x^2+1}dx=\frac{(\arctan x)^3}{3}+c$$

Work Step by Step

To evaluate the integral $$\int \frac{(\arctan x)^2}{x^2+1}dx$$ we will use substitution $\arctan x=t$ which gives us $\frac{dx}{x^2+1}=dt$. Putting this into the integral we get: $$\int\frac{(\arctan x)^2}{x^2+1}dx=\int t^2dt=\frac{t^3}{3}+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int\frac{(\arctan x)^2}{x^2+1}dx=\frac{t^3}{3}+c=\frac{(\arctan x)^3}{3}+c$$
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