Arc length $y = \frac{1}{4} x^2 - \frac{1}{2} \ln x$ $$y = \frac{1}{4} x^2 - \frac{1}{2} \ln x$$
$$\int_1^{2e} \sqrt{1 + (y')^2}$$
$$y' = \frac{x}{2} - \frac{1}{2x}$$
$$y' = \frac{2x^2-1}{2x}$$
$$\left(\frac{2x^2-1}{2x}\right)^2$$
$$\frac{4x^4-4x^2+1}{4x^2}$$
$$\int_1^{2e} \sqrt{1 + \frac{4x^4-4x^2+1}{4x^2} }$$
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?
 A: $$1+y'^2=1+\frac14\left(x^2-2+\frac1{x^2}\right)=\frac14\left(x+\frac1x\right)^2\implies$$
$$\int\limits_1^{2e}\sqrt{1+y'^2}dx=\frac12\int\limits_1^{2e}\left(x+\frac1x\right)dx=\ldots$$
A: $$y' = \frac{x}{2} - \frac{1}{2x}$$
The only problem I can see in your work is the fact that you slipped when finding a common denominator for the fractions above. Doing so correctly gives us:
$$y' = \frac{x^2-1}{2x};\quad (y')^2 = \dfrac{x^4 - 2x^2 + 1}{4x^2}$$
Now, back to the integral:
$$\begin{align} \int_1^{2e} \sqrt{1 + (y')^2} \,dx & = \int_1^{2e} \sqrt{1 +  \frac{x^4 - 2x^2 + 1}{4x^2} }\,dx \\ \\ 
& = \int_1^{2e} \sqrt{\frac{x^4 + 2x^2 + 1}{4x^2} }\,dx \\ \\
& = \int_1^{2e} \sqrt{\frac{x^2 + 2 +\frac 1{x^2}}{4} }\,dx\\ \\ 
& =  \int_1^{2e} \sqrt{\dfrac{1}{4}\left(x^2 + 2 +\frac 1{x^2}\right) }\,dx\\ \\ 
& = \int_1^{2e} \frac 12\sqrt{(x + \dfrac 1x)^2}\,dx\tag{factored}\\ \\ 
& = \dfrac 12\int_1^{2e} \left(x + \dfrac 1x\right)\,dx \\ \\ 
& = \left.\dfrac 12\left(\frac{x^2}{2} + \ln x\right)\right|_1^{2e}\end{align}$$
