3. The integer solutions of the equation $\frac{x+3}{x+1}-y=0$ are $(\quad)$ groups.
(A) 1
(B) 2
(C) 3
(D) 4
3.D.

From the problem, we know that $y=\frac{x+3}{x+1}=1+\frac{2}{x+1}$ is an integer. Therefore, $x+1= \pm 1$ or $x+1= \pm 2$. Solving this, we get $x=0,-2,1,-3$. Hence, the original equation has 4 integer solutions.