Answer
$\frac{3}{4} + \frac{1}{2} \ln 2$
Work Step by Step
$y = \frac{1}{4} x^{2} - \frac{1}{2} \ln x$ then $y' = \frac{1}{2} x - \frac{1}{2x}$ and
$1+(y')^{2} = 1+(\frac{1}{4} x^{2} -\frac{1}{2} + \frac{1}{4x^{2}}) = \frac{1}{4} x^{2} + \frac{1}{2} + \frac{1}{4x^{2}} = (\frac{1}{2} x + \frac{1}{2x})^{2}$
So
$L = \int ^{2}_{1} \sqrt{1+(y')^{2}}dx = \int ^{2}_{1} |\frac{1}{2}x + \frac{1}{2x}| dx = \int ^{2}_{1} (\frac{1}{2}x + \frac{1}{2x})dx = [\frac{1}{4}x^{2} + \frac{1}{2} \ln|x|]^{2}_{1} = \frac{3}{4} + \frac{1}{2} \ln 2$
