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: 36

Answer

$$\int\frac{16x^3}{4x^2-4x+1}dx=2x^2+4x+3\ln|2x-1|-\frac{1}{2x-1}+C$$

Work Step by Step

$$I=\int\frac{16x^3}{4x^2-4x+1}dx$$ 1) Perform long division: We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{16x^3}{4x^2-4x+1}=4x+4+\frac{12x-4}{4x^2-4x+1}=4x+4+\frac{12x-4}{(2x-1)^2}$$ 2) Express the remainder fraction as a sum of partial fractions: $$\frac{12x-4}{(2x-1)^2}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}$$ Clear fractions: $$A(2x-1)+B=12x-4$$ $$2Ax-A+B=12x-4$$ Equating coefficients of corresponding powers of $x$, we get $2A=12$, so $A=6$ $-A+B=-4$, so $B=-4+A=2$ Therefore, $$\frac{12x-4}{(2x-1)^2}=\frac{6}{2x-1}+\frac{2}{(2x-1)^2}$$ 2) Evaluate the integral: $$I=\int(4x+4)dx+\int\frac{6}{2x-1}dx+\int\frac{2}{(2x-1)^2}dx$$ $$I=2x^2+4x+\int\frac{3}{2x-1}d(2x-1)+\int\frac{1}{(2x-1)^2}d(2x-1)$$ $$I=2x^2+4x+3\ln|2x-1|-\frac{1}{2x-1}+C$$
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