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).
