Let $$c(u_1,u_2) = 1 + \alpha(1- 2u_1)(2- 2u_2)$$ where $u_1, u_2 \in (0,1)$.
The question is, for which $\alpha$ is the function $c$ a copula density for $(U_1, U_2)$.
So my idea was to compute the integral of $c$ on $[0,1]^2$, and see for which $\alpha$ is the integral equal to 1. But $$\int_0^1\int_0^1 c(u_1,u_2)\ du_1 du_2 = 1$$ for any $\alpha \in \mathbb{R}$. The answer is apparently $\alpha \in [-1,-1]$, why is that?
There must be a mistake in your question formulation because what you present cannot be a copula. For one, a copula should satisfy
$$u_1=C(u_1,1)=\int_0^1 c(u_1,u_2)du_2 = 1+\alpha(1-2u_1)$$
But this can not be satisfied for any value of $\alpha$.
I suspect what you meant was to check if
$$c(u_1,u_2) = c(u_1,u_2) = 1 + \alpha(1- 2u_1)(1- 2u_2)$$
is a copula. Note the difference in the second factor of the second term.
In that case, checking that this is a copula density corresponds to checking the conditions of the definition. Note that the first two conditions are trivial to check. The third condition is equivalent to checking that the density is positive. Now, $-1 \leq (1-2u_1)(1-2u_2) \leq 1$, thus the density can only be positive if $\alpha \in [-1,1]$.