Answer
$$e^2-\frac{1}{4}+\frac{1}{2}\ln \left(2e\right)$$
Work Step by Step
Given $$ y= \frac{1}{4}x^2-\frac{1}{2}\ln x$$
Then arc length is given by
\begin{aligned} s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x &=\int_{1}^{2 e} \sqrt{1+\left(\frac{1}{2} x-\frac{1}{2 x}\right)^{2}} d x \\ &=\int_{1}^{2 e} \sqrt{1+\frac{1}{4} x^{2}-\frac{1}{2}+\frac{1}{4 x^{2}}} d x \\ &=\int_{1}^{2 e} \sqrt{\frac{1}{2}+\frac{1}{4} x^{2}+\frac{1}{4 x^{2}}} d x \\ &=\frac{1}{2} \int_{1}^{2 e} \frac{x^{2}+1}{x} d x \\ &=\frac{1}{2} \int_{1}^{2 e}\left(x+\frac{1}{x}\right) d x \\ &=\frac{1}{2}\left[\frac{x^{2}}{2}+\ln |x|\right]_{1}^{2 e}\\
&=e^2-\frac{1}{4}+\frac{1}{2}\ln \left(2e\right) \end{aligned}
