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

Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Add $ \dfrac{3x}{x^2+2} $ and $ \dfrac{2x+1}{x^2-4} $ to get $ \dfrac{ \color{purple}{ 5x^3+x^2-8x+2 } }{ x^4-2x^2-8 }$.

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^2-4 }$ and the second by $\color{blue}{ x^2+2 }$.

$$ \begin{aligned} \frac{3x}{x^2+2} + \frac{2x+1}{x^2-4} & = \frac{ 3x \cdot \color{blue}{ \left( x^2-4 \right) }}{ \left( x^2+2 \right) \cdot \color{blue}{ \left( x^2-4 \right) }} + \frac{ \left( 2x+1 \right) \cdot \color{blue}{ \left( x^2+2 \right) }}{ \left( x^2-4 \right) \cdot \color{blue}{ \left( x^2+2 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3x^3-12x } }{ x^4-4x^2+2x^2-8 } + \frac{ \color{purple}{ 2x^3+4x+x^2+2 } }{ x^4-4x^2+2x^2-8 } = \\[1ex] &=\frac{ \color{purple}{ 5x^3+x^2-8x+2 } }{ x^4-2x^2-8 } \end{aligned} $$

Subtract $ \dfrac{2}{x} $ from $ \dfrac{5x^3+x^2-8x+2}{x^4-2x^2-8} $ to get $ \dfrac{ \color{purple}{ 3x^4+x^3-4x^2+2x+16 } }{ x^5-2x^3-8x }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x }$ and the second by $\color{blue}{ x^4-2x^2-8 }$.

$$ \begin{aligned} \frac{5x^3+x^2-8x+2}{x^4-2x^2-8} - \frac{2}{x} & = \frac{ \left( 5x^3+x^2-8x+2 \right) \cdot \color{blue}{ x }}{ \left( x^4-2x^2-8 \right) \cdot \color{blue}{ x }} - \frac{ 2 \cdot \color{blue}{ \left( x^4-2x^2-8 \right) }}{ x \cdot \color{blue}{ \left( x^4-2x^2-8 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 5x^4+x^3-8x^2+2x } }{ x^5-2x^3-8x } - \frac{ \color{purple}{ 2x^4-4x^2-16 } }{ x^5-2x^3-8x } = \\[1ex] &=\frac{ \color{purple}{ 5x^4+x^3-8x^2+2x - \left( 2x^4-4x^2-16 \right) } }{ x^5-2x^3-8x }=\frac{ \color{purple}{ 3x^4+x^3-4x^2+2x+16 } }{ x^5-2x^3-8x } \end{aligned} $$