Answer
$$\sqrt {{x^2} - 6x + 5} + 3\ln \left| {x - 3 + \sqrt {{x^2} - 6x + 5} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {{x^2} - 6x + 5} }}} dx \cr
& {\text{Completing the square}} \cr
& \int {\frac{x}{{\sqrt {{x^2} - 6x + 5} }}} dx = \int {\frac{x}{{\sqrt {{x^2} - 6x + 9 - 4} }}} dx \cr
& = \int {\frac{x}{{\sqrt {\left( {{x^2} - 6x + 9} \right) - 4} }}} dx = \int {\frac{x}{{\sqrt {{{\left( {x - 3} \right)}^2} - 4} }}} dx \cr
& {\text{Let }}u = x - 3,{\text{ }}x = u + 3,{\text{ }}dx = du \cr
& \int {\frac{x}{{\sqrt {{{\left( {x - 3} \right)}^2} - 4} }}} dx = \int {\frac{{u + 3}}{{\sqrt {{u^2} - 4} }}} du \cr
& {\text{Distribute the numerator}} \cr
& = \int {\frac{u}{{\sqrt {{u^2} - 4} }}} du + \int {\frac{3}{{\sqrt {{u^2} - 4} }}} du \cr
& = \frac{1}{2}\int {\frac{{2u}}{{\sqrt {{u^2} - 4} }}} du + 3\int {\frac{1}{{\sqrt {{u^2} - {{\left( 2 \right)}^2}} }}} du{\text{ }} \cr
& = \sqrt {{u^2} - 4} + 3\int {\frac{1}{{\sqrt {{u^2} - {{\left( 2 \right)}^2}} }}} du{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& *{\text{Integrating }}3\int {\frac{1}{{\sqrt {{u^2} - {{\left( 2 \right)}^2}} }}} du{\text{ }} \cr
& {\text{Let }}u = 2\sec \theta ,{\text{ }}du = 2\sec \theta \tan \theta d\theta \cr
& 3\int {\frac{1}{{\sqrt {{u^2} - {{\left( 2 \right)}^2}} }}} du = 3\int {\frac{{2\sec \theta \tan \theta }}{{\sqrt {4{{\sec }^2}\theta - 4} }}} d\theta {\text{ }} \cr
& = 3\int {\frac{{2\sec \theta \tan \theta }}{{\sqrt {4\left( {{{\sec }^2}\theta - 1} \right)} }}} d\theta = 3\int {\frac{{2\sec \theta \tan \theta }}{{\sqrt {4{{\tan }^2}\theta } }}} d\theta {\text{ }} \cr
& = 3\int {\frac{{2\sec \theta \tan \theta }}{{2\tan \theta }}} d\theta {\text{ }} \cr
& = 3\int {\sec \theta } d\theta {\text{ }} \cr
& {\text{Integrating}} \cr
& {\text{ = 3ln}}\left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{Where, }}\sec \theta = \frac{u}{2}{\text{ and tan}}\theta = \frac{{\sqrt {{u^2} - {{\left( 2 \right)}^2}} }}{2} \cr
& = 3\ln \left| {\frac{u}{2} + \frac{{\sqrt {{u^2} - {{\left( 2 \right)}^2}} }}{2}} \right| + C \cr
& = 3\ln \left| {u + \sqrt {{u^2} - {{\left( 2 \right)}^2}} } \right| + C \cr
& {\text{Substitute into }}\left( {\bf{1}} \right) \cr
& = \sqrt {{u^2} - 4} + 3\ln \left| {u + \sqrt {{u^2} - {{\left( 2 \right)}^2}} } \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \sqrt {{{\left( {x - 3} \right)}^2} - 4} + 3\ln \left| {x - 3 + \sqrt {{{\left( {x - 3} \right)}^2} - 4} } \right| + C \cr
& = \sqrt {{x^2} - 6x + 5} + 3\ln \left| {x - 3 + \sqrt {{x^2} - 6x + 5} } \right| + C \cr} $$
