The given equation expands to \[(a^2 + b^2) x^2 - (4ab + 1) x + a^2 + b^2 = 0.\]Since the quadratic has an integer root, its discriminant is nonnegative: \[(4ab + 1)^2 - 4(a^2 + b^2)^2 \ge 0.\]This factors as \[(4ab + 1 + 2a^2 + 2b^2)(4ab + 1 - 2a^2 - 2b^2) \ge 0.\]We can write this as \[[1 + 2(a + b)^2][1 - 2(a - b)^2] \ge 0.\]Since $ 1 + 2(a + b)^2 $ is always nonnegative, \[1 - 2(a - b)^2 \ge 0,\]so $ (a - b)^2 \le \frac{1}{2} $. Recall that $ a $ and $ b $ are integers. If $ a $ and $ b $ are distinct, then $ (a - b)^2 \ge 1, $ so we must have $ a = b $. Then the given equation becomes \[2a^2 x^2 - (4a^2 + 1) x + 2a^2 = 0.\]Let $ r $ and $ s $ be the roots, where $ r $ is the integer. Then by Vieta's formulas, \[r + s = \frac{4a^2 + 1}{2a^2} = 2 + \frac{1}{2a^2},\]and $ rs = 1 $. Since $ rs = 1, $ either both $ r $ and $ s $ are positive, or both $ r $ and $ s $ are negative. Since $ r + s $ is positive, $ r $ and $ s $ are positive. Since $ a $ is an integer, \[r + s = 2 + \frac{1}{2a^2} < 3,\]so the integer $ r $ must be 1 or 2. If $ r = 1, $ then $ s = 1, $ so both roots are integers, contradiction. Hence, $ r = 2, $ and $ s = \boxed{\frac{1}{2}} $. (For these values, we can take $ a = 1 $.)