Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 10

The integral is improper because the function under the integral f(x)=1/(x-3)^(3/2) is not defined at 3 (for x=3 denominator is equal to zero).
int_3^4 1/(x-3)^(3/2)dx=
Substitute u=x-3 => du=dx, u_l=3-3=0, u_u=4-3=1.
u_l and u_u denote new lower and upper bound respectively.
int_0^1 1/u^(3/2)du=int_0^1 u^(-3/2)du=-2u^(-1/2)|_0^1=-2cdot1^(-1/2)+lim_(u to 0)2u^(-1/2)=
-2+infty=infty
As we can see the integral diverges.
The image below shows the graph of the function and area under it corresponding to the integral. Both axis are asymptotes of the function.