## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises - Page 549: 37

#### Answer

$= - \ln \left| {{x^4}} \right| + \ln \left| {\,{{\left( {x - 1} \right)}^5}} \right| + \frac{3}{{x - 1}} + C$

#### Work Step by Step

$\begin{gathered} \int_{}^{} {\frac{{{x^2} - 4}}{{{x^3} - 2{x^2} + x}}} dx \hfill \\ \hfill \\ {\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ \frac{{{x^2} - 4}}{{{x^3} - 2{x^2} + x}} = \frac{{{x^2} - 4}}{{x\left( {{x^2} - 2x + 1} \right)}} = \frac{{{x^2} - 4}}{{x{{\left( {x - 1} \right)}^2}}} \hfill \\ \hfill \\ Applying\,\,partial\,fractions \hfill \\ \hfill \\ \frac{{{x^2} - 4}}{{x{{\left( {x - 1} \right)}^2}}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x - 1} \right)}^2}}} \hfill \\ \hfill \\ find\,\,the\,\,consta\,nts \hfill \\ \hfill \\ A = - 4,{\text{ B}} = 5,{\text{ C}} = - 3 \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{{x^2} - 4}}{{{x^3} - 2{x^2} + x}}} dx = \int_{}^{} {\,\left( { - \frac{4}{x} + \frac{5}{{x - 1}} - \frac{3}{{\,{{\left( {x - 1} \right)}^2}}}} \right)} dx \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ = - 4\ln + 5\ln \left| {x - 1} \right| + \frac{3}{{x - 1}} + C \hfill \\ \hfill \\ = - \ln \left| {{x^4}} \right| + \ln \left| {\,{{\left( {x - 1} \right)}^5}} \right| + \frac{3}{{x - 1}} + C \hfill \\ \hfill \\ \end{gathered}$

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