Answer
\[ = 8 + 3\ln 3\]
Work Step by Step
\[\begin{gathered}
\int_2^4 {\frac{{{x^2} + 2}}{{x - 1}}dx} \hfill \\
\hfill \\
use\,\,the\,\,long\,\,\,division \hfill \\
\hfill \\
= \int_2^4 {\,\left( {x + \frac{{x + 2}}{{x - 1}}} \right)dx = \int_2^4 {\,\left( {x + \frac{{x - 1 + 3}}{{x - 1}}} \right)dx} } \hfill \\
\hfill \\
= \int_2^4 {\,\left( {x + 1 + \frac{3}{{x - 1}}} \right)dx} \hfill \\
\hfill \\
integrate\,and\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= \,\,\left[ {\frac{{{x^2}}}{2} + x + 3\ln \,\left( {x - 1} \right)} \right]_2^4 \hfill \\
\hfill \\
= 8 + 4 + 3\ln 3 - \,\left( {2 + 2 + 3\ln 1} \right) \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
= 8 + 3\ln 3 \hfill \\
\end{gathered} \]