jmc

We can factor $\frac{1}{x^2+x}=\frac{1}{x(x+1)}$. Therefore, we want both $x$ and $x+1$ to be divisible by 2 and 5. They cannot both be even, so we have that either $x$ or $x+1$ is odd, so either $x$ or $x+1$ is a power of 5. We start by considering the power $5^0=1$. If $x=1$, we have the fraction $\frac{1}{2}$, which terminates. If $x+1=1$, we get $x=0$, but then we have an invalid fraction. We now consider the power $5^1=5$. If $x=5$, we have the fraction $\frac{1}{30}$, which repeats due to the factor of 3 in the denominator. If $x+1=5$, then $x=4$, so the fraction is $\frac{1}{20}=0.05$. The second smallest integer $x$ such that $\frac{1}{x^2+x}$ is a terminating decimal is $x=\boxed{4}$.