## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 446: 33

#### Answer

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

#### Work Step by Step

$$I=\int\frac{2x^3-2x^2+1}{x^2-x}dx$$ 1) Perform long division: We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{2x^3-2x^2+1}{x^2-x}=2x+\frac{1}{x^2-x}$$ 2) Express the remainder fraction as a sum of partial fractions: $$\frac{1}{x^2-x}=\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$ Clear fractions: $$Ax-A+Bx=1$$ $$(A+B)x-A=1$$ Equating coefficients of corresponding powers of $x$, we get $-A=1$, so $A=-1$ $A+B=0$, so $B=1$ Therefore, $$\frac{1}{x^2-x}=\frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}$$ 2) Evaluate the integral: $$I=\int2xdx-\int\frac{1}{x}dx+\int\frac{1}{x-1}x$$ $$I=x^2-\ln|x|+\ln|x-1|+C$$

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