Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum 
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?

As an alternative approach, since by computing residues we have
$$ \frac{3n+7}{n(n+1)(n+2)} = \frac{7}{2}\cdot\frac{1}{n}-4\cdot\frac{1}{n+1}+\frac{1}{2}\cdot\frac{1}{n+2} \tag{1}$$
it follows that
$$ S=\sum_{n\geq 1}\frac{3n+7}{n(n+1)(n+2)} = \int_{0}^{1}\left(\frac{7}{2}-4x+\frac{x^2}{2}\right)\sum_{n\geq 1}x^{n-1}\,dx \tag{2}$$
hence:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{7-8x+x^2}{1-x}\,dx = \frac{1}{2}\int_{0}^{1}(7-x)\,dx = \frac{1}{2}\left(7-\frac{1}{2}\right)=\color{red}{\frac{13}{4}}.\tag{3} $$
