Answer
$$\frac{{3x - 10}}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{3}{2}{\tan ^{ - 1}}\left( {x - 3} \right) + C{\text{ }}$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{{\left( {{x^2} - 6x + 10} \right)}^2}}}} dx \cr
& {\text{Completing the square }} \cr
& {x^2} - 6x + 9 + 1 = {\left( {x - 3} \right)^2} + 1 \cr
& \int {\frac{x}{{{{\left( {{x^2} - 6x + 10} \right)}^2}}}} dx = \int {\frac{x}{{{{\left( {{{\left( {x - 3} \right)}^2} + 1} \right)}^2}}}} dx \cr
& {\text{Let }}u = x - 3,{\text{ }}x = u + 3,{\text{ }}du = dx,{\text{ }} \cr
& \int {\frac{x}{{{{\left( {{{\left( {x - 3} \right)}^2} + 1} \right)}^2}}}} dx = \int {\frac{{u + 3}}{{{{\left( {{u^2} + 1} \right)}^2}}}} du \cr
& = \int {\frac{u}{{{{\left( {{u^2} + 1} \right)}^2}}}} du + \int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du \cr
& = \frac{1}{2}\int {\frac{{2u}}{{{{\left( {{u^2} + 1} \right)}^2}}}} du + \int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du \cr
& = \frac{1}{2}\left( { - \frac{1}{{{u^2} + 1}}} \right) + \int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du \cr
& = - \frac{1}{{2\left( {{u^2} + 1} \right)}} + \int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate }}\int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du{\text{ by tables}} \cr
& \int {\frac{{du}}{{{{\left( {{u^2} + {a^2}} \right)}^n}}} = } \frac{1}{{2{a^2}\left( {n - 1} \right)}}\left[ {\frac{u}{{{{\left( {{u^2} + {a^2}} \right)}^{n - 1}}}} + \left( {2n - 3} \right)\int {\frac{1}{{{{\left( {{u^2} + {a^2}} \right)}^{n - 1}}}}du} } \right] \cr
& {\text{For }}n = 2 \cr
& \int {\frac{{du}}{{{{\left( {{u^2} + {a^2}} \right)}^2}}} = } \frac{1}{{2{a^2}}}\left[ {\frac{u}{{\left( {{u^2} + {a^2}} \right)}} + \int {\frac{1}{{{u^2} + {a^2}}}du} } \right] \cr
& \int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du = \frac{3}{2}\left[ {\frac{u}{{{u^2} + 1}} + \int {\frac{1}{{{u^2} + 1}}du} } \right] \cr
& = \frac{{3u}}{{2\left( {{u^2} + 1} \right)}} + \frac{3}{2}{\tan ^{ - 1}}u + C \cr
& {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr
& = - \frac{1}{{2\left( {{u^2} + 1} \right)}} + \int {\frac{3}{{{{\left( {{u^2} + 1} \right)}^2}}}} du{\text{ }} \cr
& {\text{Write in terms of }}x,{\text{ }}u = x - 3 \cr
& = - \frac{1}{{2\left( {{{\left( {x - 3} \right)}^2} + 1} \right)}} + \frac{{3\left( {x - 3} \right)}}{{2\left( {{{\left( {x - 3} \right)}^2} + 1} \right)}} + \frac{3}{2}{\tan ^{ - 1}}\left( {x - 3} \right) + C \cr
& = - \frac{1}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{{3\left( {x - 3} \right)}}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{3}{2}{\tan ^{ - 1}}\left( {x - 3} \right) + C{\text{ }} \cr
& = - \frac{1}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{{3x - 9}}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{3}{2}{\tan ^{ - 1}}\left( {x - 3} \right) + C{\text{ }} \cr
& = \frac{{ - 1 + 3x - 9}}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{3}{2}{\tan ^{ - 1}}\left( {x - 3} \right) + C{\text{ }} \cr
& = \frac{{3x - 10}}{{2\left( {{x^2} - 6x + 10} \right)}} + \frac{3}{2}{\tan ^{ - 1}}\left( {x - 3} \right) + C{\text{ }} \cr} $$
