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 445: 24



Work Step by Step

$$A=\int\frac{8x^2+8x+2}{(4x^2+1)^2}dx$$ 1) Express the integrand as a sum of partial fractions: Here we have both an irreducible quadratic factor and a repeated linear factor, $$\frac{8x^2+8x+2}{(4x^2+1)^2}=\frac{Ax+B}{4x^2+1}+\frac{Cx+D}{(4x^2+1)^2}$$ Clear fractions: $$(Ax+B)(4x^2+1)+Cx+D=8x^2+8x+2$$ $$4Ax^3+Ax+4Bx^2+B+Cx+D=8x^2+8x+2$$ $$4Ax^3+4Bx^2+x(A+C)+B+D=8x^2+8x+2$$ Equating coefficients of corresponding powers of $y$, we get $A=0$ and $B=2$ $A+C=8$, so $C=8$ $B+D=2$, so $D=0$ Therefore, $$\frac{8x^2+8x+2}{(4x^2+1)^2}=\frac{2}{4x^2+1}+\frac{8x}{(4x^2+1)^2}$$ 2) Evaluate the integral: $$A=2\int\frac{dx}{4x^2+1}+\int\frac{8x}{(4x^2+1)^2}dx$$ $$A=\int\frac{d(2x)}{(2x)^2+1}+\int\frac{d(4x^2)}{(4x^2+1)^2}$$ $$A=\tan^{-1}(2x)-\frac{1}{4x^2+1}+C$$
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