<s>[INST] 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?
 [/INST] $$1+y'^2=1+\frac14\left(x^2-2+\frac1{x^2}\right)=\frac14\left(x+\frac1x\right)^2\implies$$
$$\int\limits_1^{2e}\sqrt{1+y'^2}dx=\frac12\int\limits_1^{2e}\left(x+\frac1x\right)dx=\ldots$$
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