Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 508: 52

Answer

$$\int \frac{dx}{x(x^{4}+1)}=\frac{1}{4}ln(\frac{x^{4}}{x^{4}+1})+C$$

Work Step by Step

$$\int \frac{dx}{x(x^{4}+1)}=\int \frac{x}{x^{2}(x^{4}+1)}dx$$ $$let\ t=x^{2},\,dt=2x\,dx$$ $$\int \frac{x}{x^{2}(x^{4}+1)}dx=\frac{1}{2}\int \frac{1}{t(t^{2}+1)}dt$$ $$=\frac{1}{2}\int \frac{1+t^{2}-t^{2}}{t(t^{2}+1)}dt=\frac{1}{2}\int (\frac{1}{t}-\frac{t}{t^{2}+1})dt$$ $$=\frac{1}{2}ln\left | t \right |-\frac{1}{4}ln(t^{2}+1)+C$$ $$=\frac{1}{2}ln( x^{2})-\frac{1}{4}ln(x^{4}+1)+C$$ $$=\frac{1}{4}ln(\frac{x^{4}}{x^{4}+1})+C$$
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