The problem is to simplify the expression $\frac{x^2-1}{5xy} \cdot \frac{x^2y}{1+x}$.

1. Problem Description

The problem is to simplify the expression $\frac{x^2-1}{5xy} \cdot \frac{x^2y}{1+x}$.

2. Solution Steps

First, factor the numerator $x^2 - 1$ using the difference of squares formula:

$x^2 - 1 = (x-1)(x+1)$.

Then rewrite the expression as:

$\frac{(x-1)(x+1)}{5xy} \cdot \frac{x^2y}{1+x}$.

Now, cancel out the common factors. Since $1+x = x+1$, we can cancel out $(x+1)$ from the numerator and the denominator. Also, we can cancel out $x$ and $y$ from the numerator and the denominator.

$\frac{(x-1)(x+1)}{5xy} \cdot \frac{x^2y}{x+1} = \frac{(x-1)}{5xy} \cdot \frac{x^2y}{1}$.

$= \frac{(x-1)x^2y}{5xy}$.

Cancel $xy$ from numerator and denominator:

$= \frac{(x-1)x}{5}$.

Distribute $x$ in the numerator:

$= \frac{x^2-x}{5}$.

3. Final Answer

The simplified expression is $\frac{x^2-x}{5}$.