Arc length $y = \frac{1}{4} x^2 - \frac{1}{2} lnx$

Arc length $y = \frac{1}{4} x^2 - \frac{1}{2} lnx$

$$y = \frac{1}{4} x^2 - \frac{1}{2} lnx$$
$$\int_1^{2e} \sqrt{1 + (y')^2}$$
$$y' = \frac{x}{2} - \frac{1}{2x}$$
$$y' = \frac{2x^2-1}{2x}$$
$$(\frac{2x^2-1}{2x})^2$$
$$\frac{4x^4-4x^2+1}{4x^2}$$
$$\int_1^{2e} \sqrt{1 + \frac{4x^4-4x^2+1}{4x^2} }$$
The 1 cancels out the negative term in the numerator
$$\int_1^{2e} \sqrt{ \frac{4x^4+1}{4x^2} }$$
So now if i have done this right I have now idea how to integrate this,
subsitution doesn't seem to help. What is the trick here?