By the way, notationally-speaking, it would be better to write this integral as int_{t}^{5}dx/((x-4)^2). (If you have a variable as a limit of integration, it's better to use a different letter as your integrand variable).
If t>4, then the integrand is continuous on the interval from t to 5 (or, for that matter, from 5 to t when t>5). An antiderivative of f(x)=1/((x-4)^2)=(x-4)^{-2} is F(x)=-(x-4)^{-1}. The Fundamental Theorem of Calculus then implies that int_{t}^{5}dx/((x-4)^2)=F(5)-F(t)=-(1)^{-1}-(-(t-4)^{-1})=-1+1/(t-4).
However, if t\leq 4, then f(x)=1/((x-4)^2) is not continuous on the interval from t to 5. The integral could still exist, however, as an improper integral. But, the following limit calculation shows that it does not exist because the integral from 4 to 5 diverges:
int_{4}^{5}1/((x-4)^2)\ dx=lim_{t->4+}(-1+1/(t-4)), but
lim_{t->4+}1/(t-4) does not exist (some people will write \lim_{t->4+}1/(t-4)=+infty because of the special way this limit fails to exist).