\(x=\dfrac{y^3}3+\dfrac1{4y}\)
\(\dfrac{\mathrm dx}{\mathrm dy}=y^2-\dfrac1{4y^2}\)
The curve's length along the interval [1,3] is
\(\displaystyle\int_1^3\sqrt{1+\left(\dfrac{\mathrm dx}{\mathrm dy}\right)^2}\,\mathrm dy=\int_1^3\sqrt{1+\left(y^2-\dfrac1{4y^2}\right)^2}\,\mathrm dy=\int_1^3\sqrt{y^4+\dfrac12+\dfrac1{16y^4}}\,\mathrm dy\)
Since
\(y^4+\dfrac12+\dfrac1{16y^4}=\left(y^2+\dfrac1{4y^2}\right)^2\)
you have
\(\displaystyle\int_1^3\left(y^2+\dfrac1{4y^2}\right)\,\mathrm dy=\dfrac{y^3}3-\dfrac1{4y}\bigg|_{y=1}^{y=3}=\dfrac{107}{12}-\dfrac1{12}=\dfrac{53}6\)
