Answer
\[ = \frac{1}{3}{\tan ^{ - 1}}\,\left( {\frac{{x - 1}}{3}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{{x^2} - 2x + 10}}} \hfill \\
\hfill \\
complete\,\,the\,\,square\,\,and\,\,factor \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{{x^2} - 2x + 1 + 9}}} \hfill \\
\hfill \\
= \int_{}^{} {\frac{{dx}}{{\,{{\left( {x - 1} \right)}^2} + 9}}} \hfill \\
\hfill \\
set\,\,\left( {x - 1} \right) = u\,\, \to \,\,dx = du \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{{x^2} - 2x + 10}}} = \int_{}^{} {\frac{{du}}{{{u^2} + 9}}} \hfill \\
\hfill \\
Usi\,ng\,\int_{}^{} {\frac{{dx}}{{{x^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{1}{3}{\tan ^{ - 1}}\frac{u}{3} + c \hfill \\
\hfill \\
substituting\,\,back\,\,u = \,\left( {x - 1} \right) \hfill \\
\hfill \\
= \frac{1}{3}{\tan ^{ - 1}}\,\left( {\frac{{x - 1}}{3}} \right) + C \hfill \\
\end{gathered} \]
