Add $ \dfrac{3}{x+1} $ and $ \dfrac{2}{x-1} $ to get $ \dfrac{ \color{purple}{ 5x-1 } }{ x^2-1 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x-1 }$ and the second by $\color{blue}{ x+1 }$.

$$ \begin{aligned} \frac{3}{x+1} + \frac{2}{x-1} & = \frac{ 3 \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+1 \right) \cdot \color{blue}{ \left( x-1 \right) }} + \frac{ 2 \cdot \color{blue}{ \left( x+1 \right) }}{ \left( x-1 \right) \cdot \color{blue}{ \left( x+1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3x-3 } }{ x^2 -\cancel{x}+ \cancel{x}-1 } + \frac{ \color{purple}{ 2x+2 } }{ x^2 -\cancel{x}+ \cancel{x}-1 } = \\[1ex] &=\frac{ \color{purple}{ 5x-1 } }{ x^2-1 } \end{aligned} $$

Multiply $ \dfrac{5x-1}{x^2-1} $ by $ \dfrac{x+1}{x-1} $ to get $ \dfrac{5x-1}{x^2-2x+1} $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5x-1}{x^2-1} \cdot \frac{x+1}{x-1} & \xlongequal{\text{Step 1}} \frac{ 5x-1 }{ \left( x-1 \right) \cdot \color{red}{ \left( x+1 \right) } } \cdot \frac{ 1 \cdot \color{red}{ \left( x+1 \right) } }{ x-1 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 5x-1 }{ x-1 } \cdot \frac{ 1 }{ x-1 } \xlongequal{\text{Step 3}} \frac{ \left( 5x-1 \right) \cdot 1 }{ \left( x-1 \right) \cdot \left( x-1 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 5x-1 }{ x^2-x-x+1 } = \frac{5x-1}{x^2-2x+1} \end{aligned} $$