Factor the denominator x2-8x+6.

We use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ to find the roots of $x^2-8x+6=0$.
Here, $a=1$, $b=-8$, $c=6$.

$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(6)}}{2(1)}$$ $$x = \frac{8 \pm \sqrt{64 - 24}}{2}$$ $$x = \frac{8 \pm \sqrt{40}}{2}$$ $$x = \frac{8 \pm 2\sqrt{10}}{2}$$ $$x = 4 \pm \sqrt{10}$$

So, the denominator can be factored as $(x - (4+\sqrt{10}))(x - (4-\sqrt{10}))$.

We write the fraction in the form:

$$\frac{2x+1}{x^2-8x+6} = \frac{A}{x - (4+\sqrt{10})} + \frac{B}{x - (4-\sqrt{10})}$$

Multiply both sides by the common denominator $(x - (4+\sqrt{10}))(x - (4-\sqrt{10}))$:

$$2x+1 = A(x - (4-\sqrt{10})) + B(x - (4+\sqrt{10}))$$

To find $A$, substitute $x = 4+\sqrt{10}$ into the equation:

$$2(4+\sqrt{10})+1 = A((4+\sqrt{10}) - (4-\sqrt{10})) + B(0)$$ $$8+2\sqrt{10}+1 = A(2\sqrt{10})$$ $$9+2\sqrt{10} = A(2\sqrt{10})$$ $$A = \frac{9+2\sqrt{10}}{2\sqrt{10}} = \frac{(9+2\sqrt{10})\sqrt{10}}{2\sqrt{10}\sqrt{10}} = \frac{9\sqrt{10}+2(10)}{2(10)} = \frac{20+9\sqrt{10}}{20}$$

To find $B$, substitute $x = 4-\sqrt{10}$ into the equation:

$$2(4-\sqrt{10})+1 = A(0) + B((4-\sqrt{10}) - (4+\sqrt{10}))$$ $$8-2\sqrt{10}+1 = B(-2\sqrt{10})$$ $$9-2\sqrt{10} = B(-2\sqrt{10})$$ $$B = \frac{9-2\sqrt{10}}{-2\sqrt{10}} = \frac{(9-2\sqrt{10})\sqrt{10}}{-2\sqrt{10}\sqrt{10}} = \frac{9\sqrt{10}-2(10)}{-2(10)} = \frac{9\sqrt{10}-20}{-20} = \frac{20-9\sqrt{10}}{20}$$

Substitute the values of $A$ and $B$ back into the partial fraction form:

$$\frac{2x+1}{x^2-8x+6} = \frac{\frac{20+9\sqrt{10}}{20}}{x - (4+\sqrt{10})} + \frac{\frac{20-9\sqrt{10}}{20}}{x - (4-\sqrt{10})}$$ $$\boxed{\frac{2x+1}{x^2-8x+6} = \frac{20+9\sqrt{10}}{20(x - 4 - \sqrt{10})} + \frac{20-9\sqrt{10}}{20(x - 4 + \sqrt{10})}}$$