If the distance between the parallel lines $3x+4y+7=0$ and $ax+y+b=0$ is $1$, the integral value of $b$ is

Since the given lines are parallel. Slopes of the lines are equal so $a=\frac{3}{4}$ and the two lines are $3x+4y+7=$and $3x+4y+4b=0$.

Distance between them is $\frac{|7-4b|}{\sqrt{3^2+4^2}}=1$
$\Rightarrow 7-4b=\pm 5\Rightarrow b=1/2$or $3$