University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 446: 37



Work Step by Step

$$I=\int\frac{y^4+y^2-1}{y^3+y}dy$$ 1) Perform long division: We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{y^4+y^2-1}{y^3+y}=y+\frac{(-1)}{y^3+y}=y+\frac{-1}{y(y^2+1)}$$ 2) Express the remainder fraction as a sum of partial fractions: $$\frac{-1}{y(y^2+1)}=\frac{A}{y}+\frac{By+C}{y^2+1}$$ (since $y^2+1$ is irreducible) Clear fractions: $$A(y^2+1)+(By+C)y=-1$$ $$Ay^2+A+By^2+Cy=-1$$ $$(A+B)y^2+Cy+A=-1$$ Equating coefficients of corresponding powers of $y$, we get $C=0$ and $A=-1$ $A+B=0$, so $B=1$ Therefore, $$\frac{-1}{y(y^2+1)}=-\frac{1}{y}+\frac{y}{y^2+1}$$ 2) Evaluate the integral: $$I=\int ydy-\int\frac{1}{y}dy+\int\frac{y}{y^2+1}dy$$ $$I=\frac{y^2}{2}-\ln|y|+\frac{1}{2}\int\frac{1}{y^2+1}d(y^2+1)$$ $$I=\frac{y^2}{2}-\ln|y|+\frac{1}{2}\ln|y^2+1|+C$$
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