Answer
$$\frac{1}{2}\ln \left| {{x^2} - 3 + \sqrt {{x^4} - 6{x^2} + 5} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {{x^4} - 6{x^2} + 5} }}} dx \cr
& {\text{Complete the square}} \cr
& \int {\frac{x}{{\sqrt {\left( {{x^4} - 6{x^2} + 5 + 4} \right) - 4} }}} dx \cr
& \int {\frac{x}{{\sqrt {\left( {{x^4} - 6{x^2} + 9} \right) - 4} }}} dx \cr
& \int {\frac{x}{{\sqrt {{{\left( {{x^2} - 3} \right)}^2} - {2^2}} }}} dx \cr
& {\text{Let }}u = {x^2} - 3,{\text{ }}du = 2xdx,{\text{ }}dx = \frac{{du}}{{2x}} \cr
& {\text{Substituting}} \cr
& \int {\frac{x}{{\sqrt {{{\left( {{x^2} - 3} \right)}^2} - {2^2}} }}} dx = \int {\frac{x}{{\sqrt {{u^2} - {2^2}} }}} \left( {\frac{{du}}{{2x}}} \right) \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt {{u^2} - {2^2}} }}} du \cr
& {\text{Integrate by tables }}\int {\frac{1}{{\sqrt {{u^2} - {a^2}} }}du} = \ln \left| {{x^4} - 6{x^2} + 5} \right| + C \cr
& \frac{1}{2}\int {\frac{1}{{\sqrt {{u^2} - {2^2}} }}} du = \frac{1}{2}\ln \left| {u + \sqrt {{u^2} - {2^2}} } \right| + C \cr
& = \frac{1}{2}\ln \left| {u + \sqrt {{u^2} - 4} } \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{2}\ln \left| {{x^2} - 3 + \sqrt {{{\left( {{x^2} - 3} \right)}^2} - 4} } \right| + C \cr
& = \frac{1}{2}\ln \left| {{x^2} - 3 + \sqrt {{x^4} - 6{x^2} + 5} } \right| + C \cr} $$
