Use the distributive property.

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

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

Multiply $ \dfrac{x^2-x+1}{x^2-1} $ by $ 7 $ to get $ \dfrac{ 7x^2-7x+7 }{ x^2-1 } $.

Step 1: Write $ 7 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^2-x+1}{x^2-1} \cdot 7 & \xlongequal{\text{Step 1}} \frac{x^2-x+1}{x^2-1} \cdot \frac{7}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2-x+1 \right) \cdot 7 }{ \left( x^2-1 \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 7x^2-7x+7 }{ x^2-1 } \end{aligned} $$