Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises - Page 549: 31

Answer

$$ - \frac{2}{x} + 2\ln \left| {\frac{{x + 1}}{x}} \right|$$

Work Step by Step

$$\eqalign{ & \int {\frac{2}{{{x^3} + {x^2}}}dx} \cr & {\text{integrand }} \cr & = \frac{2}{{{x^3} + {x^2}}} \cr & {\text{factor}} \cr & = \frac{2}{{{x^2}\left( {x + 1} \right)}} \cr & {\text{partial fractions}} \cr & \frac{2}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} \cr & 2 = Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C{x^2} \cr & {\text{multiplying}} \cr & 2 = A{x^2} + Ax + Bx + B + C{x^2} \cr & 2 = \left( {A{x^2} + C{x^2}} \right) + \left( {Ax + Bx} \right) + B \cr & {\text{by equating the coefficients}} \cr & {x^2}:{\text{ }}A + C = 0 \cr & x:{\text{ }}A + B = 0 \cr & {x^0}:{\text{ }}B = 2 \cr & {\text{Solving these equations}} \cr & A = - 2 \cr & B = 2 \cr & C = 2 \cr & {\text{substituting constants}} \cr & \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} = \frac{{ - 2}}{x} + \frac{2}{{{x^2}}} + \frac{2}{{x + 1}} \cr & \int {\frac{2}{{{x^3} + {x^2}}}dx} = \int {\left( {\frac{{ - 2}}{x} + \frac{2}{{{x^2}}} + \frac{2}{{x + 1}}} \right)dx} \cr & {\text{integrating}} \cr & = - 2\ln \left| x \right| - \frac{2}{x} + 2\ln \left| {x + 1} \right| \cr & {\text{Simplify}} \cr & = - \frac{2}{x} + 2\ln \left| {\frac{{x + 1}}{x}} \right| \cr} $$
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