Answer
\[ = \frac{{{x^2}}}{2} + \frac{1}{9}\ln \left| x \right| - \frac{{41}}{9}\ln \left( {{x^2} + 9} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{x^4} + 1}}{{{x^3} + 9x}}\,} dx \hfill \\
\hfill \\
Given\,\,function \hfill \\
\hfill \\
\frac{{{x^4} + 1}}{{{x^3} + 9x}} \hfill \\
\hfill \\
performing\,\,long\,\,division \hfill \\
\hfill \\
\frac{{{x^4} + 1}}{{{x^3} + 9x}} = x + \frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} \hfill \\
\hfill \\
Using\,\,partial\,\,fraction\,\,for\,\,\,\frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} \hfill \\
\hfill \\
\frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} = \frac{A}{{9x}} + \frac{{Bx + C}}{{9\,\left( {{x^2} + 9} \right)}} \hfill \\
\hfill \\
{\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\
A = 1,{\text{ B}} = - 82/9{\text{ and }}B = 0 \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\frac{{1 - 9{x^2}}}{{x\,\left( {{x^2} + 9} \right)}} = \frac{1}{{9x}} - \frac{{82x}}{{9\,\left( {{x^2} + 9} \right)}} \hfill \\
\hfill \\
\frac{{{x^4} + 1}}{{{x^3} + 9x}} = x + \frac{1}{{9x}} - \frac{{82x}}{{9\,\left( {{x^2} + 9} \right)}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{{x^4} + 1}}{{{x^3} + 9x}}\,} dx = \int_{}^{} {\,\left( {x + \frac{1}{{9x}} - \frac{{82x}}{{9\,\left( {{x^2} + 9} \right)}}} \right)\,dx} \hfill \\
\hfill \\
\textbf{ integrate} \hfill \\
\hfill \\
=\boxed{ \frac{{{x^2}}}{2} + \frac{1}{9}\ln \left| x \right| - \frac{{41}}{9}\ln \left( {{x^2} + 9} \right) + C} \hfill \\
\end{gathered} \]
