Answer
\[ = \ln {\left| {\frac{{\,\left( {x + 1} \right)\,{{\left( {x - 3} \right)}^{\frac{1}{3}}}}}{{\,\left( {x - 1} \right)\,{{\left( {x + 3} \right)}^{\frac{1}{3}}}}}} \right|^{\frac{1}{{16}}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{{x^4} - 10{x^2} + 9}}} \hfill \\
\hfill \\
{\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
\frac{1}{{{x^4} - 10{x^2} + 9}} = \frac{1}{{\,\left( {x - 3} \right)\,\left( {x + 3} \right)\,\left( {x - 1} \right)\,\left( {x + 1} \right)}} \hfill \\
\hfill \\
Using\,\,partial\,fractions \hfill \\
\hfill \\
\frac{1}{{\,\left( {x - 3} \right)\,\left( {x + 3} \right)\,\left( {x - 1} \right)\,\left( {x + 1} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{x + 3}} + \frac{C}{{x - 1}} + \frac{D}{{x + 1}} \hfill \\
\hfill \\
{\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\
\hfill \\
A = - \frac{1}{{48}},\,\,B = \frac{1}{{48}},\,\,C = - \frac{1}{{16}},\,\,D = \frac{1}{{16}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{1}{{\,\left( {x - 3} \right)\,\left( {x + 3} \right)\,\left( {x - 1} \right)\,\left( {x + 1} \right)}} = \frac{{ - \frac{1}{{48}}}}{{x + 3}} + \frac{{\frac{1}{{48}}}}{{x - 3}} + \frac{{ - \frac{1}{{16}}}}{{x - 1}} + \frac{{\frac{1}{{16}}}}{{x + 1}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{{x^4} - 10{x^2} + 9}}} = \int_{}^{} {\,\left( {\frac{{ - \frac{1}{{16}}}}{{x - 1}} + \frac{{\frac{1}{{16}}}}{{x + 1}} + \frac{{ - \frac{1}{{48}}}}{{x + 3}} + \frac{{\frac{1}{{48}}}}{{x - 3}}} \right)} \hfill \\
\hfill \\
{\text{integrating}} \hfill \\
\hfill \\
= - \frac{1}{{16}}\ln \left| {x - 1} \right| + \frac{1}{{16}}\ln \left| {x + 1} \right| - \frac{1}{{48}}\ln \left| {x + 3} \right| + \frac{1}{{48}}\ln \left| {x - 3} \right| + C \hfill \\
\hfill \\
= \ln {\left| {\frac{{\,\left( {x + 1} \right)\,{{\left( {x - 3} \right)}^{\frac{1}{3}}}}}{{\,\left( {x - 1} \right)\,{{\left( {x + 3} \right)}^{\frac{1}{3}}}}}} \right|^{\frac{1}{{16}}}} + C \hfill \\
\hfill \\
\end{gathered} \]
