Simplify the first fraction:
[tex] \frac{3n^{2}-n}{n^{2}-1} = \frac{3n(n-1)}{(n+1)(n-1)} = \frac{3n}{n+1} [/tex]
Your equation becomes:
[tex] \frac{3n}{n+1} + \frac{n^{2} }{n+1} \\ = \frac{n^{2} + 3n}{n+1} \\ = \frac{n(n + 3)}{n+1} [/tex]
That last fraction becomes your final answer.
