How to calculate F dot dr

The problem is:
Let C be the curve parameterized by x(t)=tan(t), y(t)=t^2, 0 <= t <= 1. For the conservative field F = (y-2x-1)i + (x+9y^2+4)j, calculate integral F dot dr. (There's also a Capital C next to the integral symbol, if that helps).
My professor put "Try the FILI" next to this on the graded test. I'm not entirely sure what that means, or maybe he was just trying to recommend me a good sandwich and didn't know how to spell "Philly" ?

F = (y-2x-1)i + (x+9y^2+4)j

P(x,y) = y-2x-1
Q(x,y) = x+9y^2+4

∂P/∂y = 1; ∂Q/∂x = 1
The field is conservative and so we know that a potential function, f(x,y) exists

Line integral of conservative fields have values that are INDEPENDENT from the path, they depend only by initial point and final point, that is

∫(F· dr) =
c

∫(∇f · dr) = f(r(1)) -f(r(0))
c

r(1) = (tan 1, 1)
r(0) = (0,0)


to find f we know that it must be
∇f = (∂f/∂x)i + (∂f/∂y)j = Pi + Qj = F

thus we have to find f such that
(∂f/∂x) = P
(∂f/∂y) = Q

f(x,y) = ∫P(x,y) dx or f(x,y) = ∫Q(x,y) dy

f(x,y) = ∫(y - 2x - 1)dx = xy - x^2 - x + h(y)

[h(y) is similar to the integration constant C of one variable functions integration:
if you differentiate wrt x then ∂h/∂x = 0]

to find the function h(y) remember that (∂f/∂y) = Q(x,y) = x + 9y^2 + 4
then differentiate f wrt y

(∂f/∂y) = x + h'(y)
and this must be Q(x,y) = x + 9y^2 + 4

so we have
x + h'(y) = x + 9y^2 + 4

that is
h'(y) = 9y^2 + 4

and integrating
h(y) = ∫(9y^2 + 4) dy = 3y^3 + 4y + C

Potential function we have found is
f(x,y) = xy - x^2 - x + 3y^3 + 4y + C
and finally

∫(F· dr) = f(tan 1, 1) - f(0,0) = tan 1 - (tan 1)^2 - tan 1 + 3 + 4 + C - C = 7 - (tan 1)^2 ≈ 4.57448
c

look at the tutorial below

Find ∫c F·dr

Let F=(y-2x-1)i + (x+9y²+4)j = P(x,y)i + Q(x,y)j
r(t) = = , 0≤t≤1
r(1) =
r(0) = = <0, 0>

Check if potential function exists.
∂P/∂y = 1
∂Q/∂x = 1

Since ∂P/∂y = ∂Q/∂x, the potentail finction exists.

Find the potential function f(x,y) such that ∇f = <(y-2x-1), (x+9y²+4)>
∂f/∂x = y-2x-1
f = yx - x² - x + h(y)
∂f/∂y = x + dh/dy = x+9y²+4
dh/dy = 9y²+4
h = 3y³ + 4y + C
Thus, f(x,y) = xy - x² - x + 3y³ + 4y + C

By FTLI,
∫c F·dr, 0≤t≤1
= f(r(1)) - f(r(0))
= tan1 - tan²1 - tan1 + 3 + 4 - 0
=7 - tan²1