Bounds for $\sum_{n\geq 2}\frac{n+2}{(n^2-1)(n+3)}$ Use the integral test to prove this inequaliy
I calculated the integral $\int_{2}^{\infty}\frac{2+x}{(x^2-1)(x+3)}dx$
How can I use the integral test to show that $0.45 < \sum_{n=2}^{\infty}\frac{2+n}{(n^2-1)(n+3)} < 0.75$ ?
I know I can use the integral test but I don't quiet understand how this helps

Why do you need the integral test? The series can be computed through partial fraction decomposition:
$$\frac{n+2}{(n-1)(n+1)(n+3)}=\frac{3}{8}\cdot\frac{1}{n-1}-\frac{1}{4}\cdot\frac{1}{n+1}-\frac{1}{8}\cdot\frac{1}{n+3}\tag{1}$$
together with:
$$ \sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) = \frac{3}{2},$$
$$ \sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n+3}\right) = \frac{25}{12}\tag{2}$$
lead to:

$$\sum_{n\geq 2}\frac{n+2}{(n-1)(n+1)(n+3)}=\color{red}{\frac{61}{96}}=0.63541666\ldots\tag{3}$$

We may notice that the value of the integral $\int_{2}^{+\infty}\frac{x+2}{(x^2-1)(x+3)}\,dx = \frac{\log(45)}{8} = 0.47583\ldots$ is just a weak lower bound and the value of the integral $\int_{3/2}^{+\infty}\frac{x+2}{(x^2-1)(x+3)}\,dx = \frac{\log(15)}{4} = 0.67701\ldots $ is just a slightly better upper bound.
