Answer
$$\frac{53}{6}$$
Work Step by Step
Given $$y=\frac{x^{3}}{3}+\frac{1}{4 x}, \quad 1 \leq x \leq 3$$
Since
\begin{align*}
y'&= x^2 -\frac{1}{4x^2} \\
y'^2&= \left( x^2 -\frac{1}{4x^2} \right)^2\\
&= x^4-\frac{1}{2} +\frac{1}{16x^4}
\end{align*}
Then
\begin{align*}
1+y'^2&= 1+x^4-\frac{1}{2} +\frac{1}{16x^4}\\
&=x^4+\frac{1}{2} +\frac{1}{16x^4}\\
&=\left(x^2+\frac{1}{4x^2} \right)^2
\end{align*}
Hence, the arc length is given by
\begin{align*}
L&=\int_{1}^{3}\sqrt{1+y'^2}dx\\
&= \int_{1}^{3}\sqrt{\left(x^2+\frac{1}{4x^2} \right)^2}dx\\
&= \int_{1}^{3} \left(x^2+\frac{1}{4x^2} \right)dx\\
&=\frac{x^3}{3} -\frac{1}{4x}\bigg|_{1}^{3}\\
&=\frac{26}{3}+\frac{1}{6}\\
&= \frac{53}{6}
\end{align*}
