To simplify the expression \(\frac{-2}{p+1} + \frac{4p}{p^2-1}\), we first note that \(p^2 - 1\) can be factored as \((p-1)(p+1)\). Thus, we rewrite the expression:
\[ \frac{-2}{p+1} + \frac{4p}{(p-1)(p+1)} \]
Next, we find a common denominator, which is \((p-1)(p+1)\).
Now, we will rewrite each fraction with this common denominator:
- The first fraction becomes:
\[ \frac{-2}{p+1} = \frac{-2(p-1)}{(p-1)(p+1)} = \frac{-2p + 2}{(p-1)(p+1)} \]
- The second fraction is already using the common denominator:
\[ \frac{4p}{(p-1)(p+1)} \]
Now, we can combine the two fractions:
\[ \frac{-2p + 2 + 4p}{(p-1)(p+1)} = \frac{2p + 2}{(p-1)(p+1)} \]
Next, we can factor the numerator:
\[ 2(p + 1) \]
Thus, the expression simplifies to:
\[ \frac{2(p + 1)}{(p-1)(p+1)} \]
We can further simplify this by cancelling \((p + 1)\) (as long as \(p \neq -1\)): \[ \frac{2}{p - 1} \]
So the simplified expression is:
\[ \frac{2}{p - 1} \quad \text{for } p \neq -1 \]
This gives us the final answer:
\[ \frac{2}{p - 1} \]