[tex]y=\dfrac13(x^2+2)^{3/2}\implies y'=x(x^2+2)^{1/2}\implies (y')^2=x^4+2x^2[/tex]
The arc length is given by
[tex]\displaystyle\int_{x=0}^{x=3}\sqrt{1+(y')^2}\,\mathrm dx=\int_0^3\sqrt{x^4+2x^2+1}\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^3\sqrt{(x^2+1)^2}\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^3(x^2+1)\,\mathrm dx[/tex]
[tex]=\left(\dfrac13x^3+x\right)\bigg|_0^3[/tex]
[tex]=12[/tex]
