By the fundamental theorem of calculus, we have
[tex]\dfrac{\mathrm dy}{\mathrm dx}=x^2+x[/tex]
[tex]\implies y(x)=y(3)+\displaystyle\int_3^x(t^2+t)\,\mathrm dt[/tex]
[tex]y(x)=17+\left(\dfrac{t^3}3+\dfrac{t^2}2\right)\bigg|_3^x[/tex]
[tex]y(x)=17+\left(\dfrac{x^3}3+\dfrac{x^2}2-\dfrac{3^3}3-\dfrac{3^2}2\right)[/tex]
[tex]y(x)=\dfrac{x^3}3+\dfrac{x^2}2+\dfrac72[/tex]
