Intuitively, #\sum_{n=1}^{\infty}1/(n^2+4)# should converge since it is "like" the #p#-series #\sum_{n=1}^{\infty}1/n^2# which converges since #p=2>1#. In fact, you can even use the comparison test directly with this series to show that #\sum_{n=1}^{\infty}1/(n^2+4)# converges.
But now on to the integral test as requested:
The function #f(x)=1/(x^2+4)# is continuous, positive, and decreasing for all #x\geq 1#. Moreover, #\int_{1}^{\infty}f(x)\ dx=\frac{1}{4}\int_{1}^{\infty}\frac{1}{1+(x/2)^2}#
#=lim_{b->\infty}(\frac{1}{2}arctan(x/2))|_{1}^{b}=\frac{1}{2}\cdot \frac{\pi}{2}-\frac{1}{2}\cdot arctan(1/2)#
In other words, the improper integral converges.
The integral test now implies that the series #\sum_{n=1}^{\infty}f(n)=\sum_{n=1}^{\infty}1/(n^2+4)# converges.
