[tex]y=\dfrac{x^6+8}{16x^2}\implies y'=\dfrac{x^6-4}{4x^3}[/tex]
The length of the curve over the given integral is
[tex]\displaystyle\int_2^3\sqrt{1+(y')^2}\,\mathrm dx[/tex]
[tex]\displaystyle\int_2^3\sqrt{1+\frac{(x^6-4)^2}{16x^6}}\,\mathrm dx[/tex]
[tex]\displaystyle\int_2^3\sqrt{1+\frac{x^{12}-8x^6+16}{16x^6}}\,\mathrm dx[/tex]
[tex]\displaystyle\int_2^3\sqrt{\frac{x^{12}+8x^6+16}{16x^6}}\,\mathrm dx[/tex]
[tex]\displaystyle\int_2^3\sqrt{\frac{(x^6+4)^2}{16x^6}}\,\mathrm dx[/tex]
[tex]\displaystyle\frac14\int_2^3\frac{x^6+4}{x^3}\,\mathrm dx[/tex]
[tex]\displaystyle\frac14\int_2^3\left(x^3+\frac4{x^3}\right)\,\mathrm dx[/tex]
[tex]\displaystyle\frac14\left(\frac{x^4}4-\frac2{x^2}\right)\bigg|_{x=2}^{x=3}=\frac{595}{144}[/tex]
