Multiply $5$ by $ \dfrac{x}{3x+3} $ to get $ \dfrac{ 5x }{ 3x+3 } $.

Step 1: Write $ 5 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Add $ \dfrac{1}{x^2+5x+4} $ and $ \dfrac{5x}{3x+3} $ to get $ \dfrac{ \color{purple}{ 5x^2+20x+3 } }{ 3x^2+15x+12 }$.

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}{ 3 }$ and the second by $\color{blue}{ x+4 }$.

$$ \begin{aligned} \frac{1}{x^2+5x+4} + \frac{5x}{3x+3} & = \frac{ 1 \cdot \color{blue}{ 3 }}{ \left( x^2+5x+4 \right) \cdot \color{blue}{ 3 }} + \frac{ 5x \cdot \color{blue}{ \left( x+4 \right) }}{ \left( 3x+3 \right) \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3 } }{ 3x^2+15x+12 } + \frac{ \color{purple}{ 5x^2+20x } }{ 3x^2+15x+12 }=\frac{ \color{purple}{ 5x^2+20x+3 } }{ 3x^2+15x+12 } \end{aligned} $$