Let [tex]\mathbf f(x,y)=\langle y^2+2x+1,2xy+4y-1\rangle[/tex]. Then
[tex]\dfrac{\partial(y^2+2x+1)}{\partial y}=2y[/tex]
[tex]\dfrac{\partial(2xy+4y-1)}{\partial x}=2y[/tex]
which means [tex]\mathbf f(x,y)[/tex] represents a conservative vector field. This means there is some scalar function [tex]f(x,y)[/tex] such that [tex]\nabla f(x,y)=\mathbf f(x,y)[/tex], and so the gradient theorem applies. This means the value of the line integral is path independent. We have
[tex]\dfrac{\partial f}{\partial x}=y^2+2x+1[/tex]
[tex]\implies f(x,y)=\displaystyle\int(y^2+2x+1)\,\mathrm dx[/tex]
[tex]f(x,y)=xy^2+x^2+x+g(y)[/tex]
[tex]\implies\dfrac{\partial f}{\partialy}=2xy+g'(y)=2xy+4y-1[/tex]
[tex]\implies g'(y)=4y-1[/tex]
[tex]\implies g(y)=\displaystyle\int(4y-1)\,\mathrm dy[/tex]
[tex]g(y)=2y^2-y+C[/tex]
and so we have
[tex]f(x,y)=xy^2+x^2+x+2y^2-y+C[/tex]
Then the value of the line integral is
[tex]\displaystyle\int_C(y^2+2x+1)\,\mathrm dx+(2xy+4y-1)\,\mathrm dy=f(3,9)-f(0,0)=408[/tex]
