Answer
doesn't exist
Work Step by Step
Given
$$\int_1^5\frac{dt}{(t-4)^2}$$
Since $f(t)=\frac{dt}{(t-4)^2} $ has an infinite discontinuity at $t = 4$;
that is, $f(t)$ is discontinuous on the interval $[1,5]$. Then $\displaystyle\int_1^5\frac{dt}{(t-4)^2}$ doesn't exist