Multiply $ \dfrac{3x+1}{x^2-1} $ by $ \dfrac{x+1}{3x^2+x} $ to get $ \dfrac{ 1 }{ x^2-x } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{3x+1}{x^2-1} \cdot \frac{x+1}{3x^2+x} & \xlongequal{\text{Step 1}} \frac{ 1 \cdot \color{blue}{ \left( 3x+1 \right) } }{ \left( x-1 \right) \cdot \color{red}{ \left( x+1 \right) } } \cdot \frac{ 1 \cdot \color{red}{ \left( x+1 \right) } }{ x \cdot \color{blue}{ \left( 3x+1 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 1 }{ x-1 } \cdot \frac{ 1 }{ x } \xlongequal{\text{Step 3}} \frac{ 1 \cdot 1 }{ \left( x-1 \right) \cdot x } \xlongequal{\text{Step 4}} \frac{ 1 }{ x^2-x } \end{aligned} $$