Square both sides (valid since both sides are non-negative):

$(2x - 5)^2 < (x + 4)^2$

$4x^2 - 20x + 25 < x^2 + 8x + 16$

$3x^2 - 28x + 9 < 0$

Find roots: $x = \frac{28 \pm \sqrt{784 - 108}}{6} = \frac{28 \pm \sqrt{676}}{6} = \frac{28 \pm 26}{6}$

$x = \frac{54}{6} = 9$ or $x = \frac{2}{6} = \frac{1}{3}$

Since coefficient of $x^2$ is positive, the parabola opens upward.

Solution: $\frac{1}{3} < x < 9$

Integer values: $1, 2, 3, 4, 5, 6, 7, 8$ → $8$ integers

Answer: B) $8$