Add $ \dfrac{6}{x^2-4x+5} $ and $ \dfrac{1}{x+1} $ to get $ \dfrac{ \color{purple}{ x^2+2x+11 } }{ x^3-3x^2+x+5 }$.

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 $\color{blue}{ x^2-4x+5 }$.

$$ \begin{aligned} \frac{6}{x^2-4x+5} + \frac{1}{x+1} & = \frac{ 6 \cdot \color{blue}{ \left( x+1 \right) }}{ \left( x^2-4x+5 \right) \cdot \color{blue}{ \left( x+1 \right) }} + \frac{ 1 \cdot \color{blue}{ \left( x^2-4x+5 \right) }}{ \left( x+1 \right) \cdot \color{blue}{ \left( x^2-4x+5 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 6x+6 } }{ x^3+x^2-4x^2-4x+5x+5 } + \frac{ \color{purple}{ x^2-4x+5 } }{ x^3+x^2-4x^2-4x+5x+5 } = \\[1ex] &=\frac{ \color{purple}{ x^2+2x+11 } }{ x^3-3x^2+x+5 } \end{aligned} $$