Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{{16}}\ln \left| {x + 2} \right| - \frac{1}{{16}}\ln \left| {x - 2} \right| - \frac{1}{{4\,{{\left( {x - 2} \right)}^2}}} + C$
$\begin{gathered} \int_{}^{} {\frac{{dx}}{{{x^3} - 2{x^2} - 4x + 8}}} \hfill \\ \hfill \\ factor\,\,the\,\,denominator \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\,\left( {x + 2} \right)\,\left( {x - 2} \right)}}} \hfill \\ \hfill \\ descomposition\,\,into\,partial\,fraction \hfill \\ \hfill \\ \frac{1}{{\,\left( {x + 2} \right)\,\left( {x - 2} \right)}} = \frac{A}{{x + 2}} + \frac{B}{{x - 2}} + \frac{C}{{\,{{\left( {x - 2} \right)}^2}}} \hfill \\ \hfill \\ find\,A,B,C \hfill \\ \hfill \\ A = \frac{1}{{16}}\,\,,\,\,B = - \frac{1}{{16}}\,\,,\,C = \frac{1}{4} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_{}^{} {\frac{1}{{16\,\left( {x + 2} \right)}}dx} - \frac{1}{{16}}\int_{}^{} {\frac{1}{{x - 2}}dx} + \frac{1}{4}\int_{}^{} {\frac{{dx}}{{\,{{\left( {x - 2} \right)}^2}}}} \hfill \\ \hfill \\ \operatorname{int} egrating \hfill \\ \hfill \\ = \frac{1}{{16}}\ln \left| {x + 2} \right| - \frac{1}{{16}}\ln \left| {x - 2} \right| - \frac{1}{{4\,{{\left( {x - 2} \right)}^2}}} + C \hfill \\ \end{gathered}$