One way to show that an improper integral such as f(x) dx converges is to compare it to another convergent integral which is easier to calculate. If 0 1 To show that 1 (5 + 26) 1/2 dx 1 converges we will compare it to dx, which we know converges since it is a p- 6/2 integral with p = 3 >1. Since r 2 1 + + 5 implies that Vx + 5 and hence Integrating both sides of this inequality over 1, oo) completes the argument