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