Use polynomial long division to rewrite the following fraction in the form $q(x)+\frac{r(x)}{d(x)}$, where $d(x)$ is the denominator of the original fraction, $q(x)$ is the quotient, and $r(x)$ is the remainder.
$$\frac{4x^3-4x^2+4x+12}{2x+2}$$

Step 1 ：We have the fraction $\frac{4x^3-4x^2+4x+12}{2x+2}$. We can rewrite this as a polynomial division problem: $4x^3-4x^2+4x+12$ divided by $2x+2$.

Step 2 ：We can simplify the divisor $2x+2$ to $x+1$ by factoring out 2.

Step 3 ：Now we perform polynomial long division. The first term of the quotient $q(x)$ is $4x^2$, because $4x^3$ divided by $x$ is $4x^2$.

Step 4 ：Subtract $4x^2(x+1)$ from $4x^3-4x^2$ to get the new dividend $-4x^2+4x+12$.

Step 5 ：The next term of the quotient is $-4x$, because $-4x^2$ divided by $x$ is $-4x$.

Step 6 ：Subtract $-4x(x+1)$ from $-4x^2+4x+12$ to get the new dividend $4x+12$.

Step 7 ：The next term of the quotient is $4$, because $4x$ divided by $x$ is $4$.

Step 8 ：Subtract $4(x+1)$ from $4x+12$ to get the new dividend $8$.

Step 9 ：Since the degree of the new dividend $8$ is less than the degree of the divisor $x+1$, we stop here. The remainder $r(x)$ is $8$.

Step 10 ：So, the original fraction can be rewritten as $q(x)+\frac{r(x)}{d(x)}$, which is $4x^2-4x+4+\frac{8}{x+1}$.

Step 11 ：Therefore, the final answer is $\boxed{{\displaystyle 4x^2-4x+4+\frac{8}{x+1}}}$.