$
f(z) = x^2 + ixy \\ $; A(1,1,) to B(2,4)
x = t; y = t$^2$; dx = dt; dy = 2t dt
dz = dx + idy
= dt + i(2t dt)
= (1+2it) dt
$ f(z) = t^2 + i(t)(t^2) \\
= t^2 + it^3 $
$
\int_A^B f(z) \,\, dx
= \int_1^2 (t^2 + it^3)(dt + i(2t) dt) \\
= \int_1^2 (t^2 + it^3)(1+2it) \,\, dt \\
= \int_0^1 (t^2 + 2it^3+it^3-2t^4) \,\, dt \\
= \int_0^1 (t^2 + 3it^3-2t^4) \,\, dt \\
= [\frac{t^3}{3} + \frac{3it^4}{4} - \frac{2t^5}{5}]_1^2 \\
= \frac{7}{3} + \frac{48i}{4} - \frac{3i}{4} - \frac{64}{5} + \frac{2}{5} \\
= \frac{7}{3} - \frac{62}{5} + \frac{45i}{4} \\
= \frac{-151}{15} + \frac{45i}{4}
$