Recall that indefinite integral follows `int f(x) dx = F(x) +C` where:

`f(x)` as the integrand function

`F(x)` as the antiderivative of `f(x)`

`C ` as the constant of integration

The given integral problem: `int x/(x^4-6x^2+5) dx` resembles one of the formulas from the integration table. We follow the integral formula for rational function with roots as:

`int (dx)/sqrt(ax^2+bx+c) = 1/sqrt(a)ln|2ax+b+2sqrt(a(ax^2+bx+c))| +C` .

For easier comparison, we apply u-substitution by letting: `u=x^2`

then `du= 2x dx` or `(du)/2 =xdx` .

Plug-in the values, we get:

`int x/(x^4-6x^2+5) dx =int 1/(x^4-6x^2+5)*x dx`

                                ` =int 1/(u^2-6u+5)*(du)/2`

Apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .

`int 1/(u^2-6u+5)*(du)/2 = 1/2int 1/(u^2-6u+5) du`

By comparing `ax^2+bx+c`  with `u^2-6u+5` , we determine the corresponding values as: `a=1` , `b=-6` ,and `c=5` .

Applying the aforementioned formula for rational function with roots, we get:

`1/2int 1/(u^2-6u+5) du`

`=1/2 * [1/sqrt(1)ln|2(1)u+(-6)+2sqrt(1(1u^2+(-6)u+5))|] +C`

`=1/2 * [1/1ln|2u-6+2sqrt(u^2-6u+5)|] +C`

`=(ln|2u-6+2sqrt(u^2-6u+5)|)/2 +C`

Plug-in `u = x^2`  and `u^2=x^4 `  on   `(ln|2u-6+2sqrt(u^2-6u+5)|)/2 +C` , we get the indefinite integral as:

`int x/(x^4-6x^2+5) dx =(ln|2x^2-6+2sqrt(x^4-6x^2+5)|)/2 +C`