You have some missing symbols, so I'm guessing that the integral reads
[tex]\displaystyle \int_C (x-9y) \, dx - x^2 \, dy[/tex]
where C is composed of the two line segments,
• C₁ = {(9t, t) : 0 ≤ t ≤ 1}
• C₂ = {(9 + t, 1 - t) : 0 ≤ t ≤ 1}
The integrals over each path are
[tex]\displaystyle \int_{C_1} (x-9y) \, dx - x^2 \, dy = \int_0^1 (9t - 9\cdot t) (9 \, dt) - (9t)^2 \, dt[/tex]
[tex]\displaystyle \cdots = -81 \int_0^1 t^2 \, dt[/tex]
[tex]\displaystyle \cdots = -\frac{81}3 (1^3 - 0^3) = -27[/tex]
and
[tex]\displaystyle \int_{C_2} (x-9y) \, dx - x^2 \, dy = \int_0^1 ((9 + t) - 9(1 - t)) \, dt - (9 + t)^2 (-dt)[/tex]
[tex]\displaystyle \cdots = \int_0^1 (81 + 28t + t^2) \, dt[/tex]
[tex]\displaystyle \cdots = 81 (1 - 0) + 14 (1^2 - 0^2) + \frac13 (1^3 - 0^3) = \frac{286}3[/tex]
Then the overall line integral has a value of
[tex]\displaystyle \int_C (x-9y) \, dx - x^2 \, dy = -27 + \frac{286}3 = \boxed{\frac{205}3}[/tex]
