## Calculus 8th Edition

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