Subtract $ \dfrac{4}{b^2-2b+1} $ from $ \dfrac{b^2+3b}{b^2-2b+1} $ to get $ \dfrac{ b^2+3b - 4 }{ \color{blue}{ b^2-2b+1 }}$.

To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{b^2+3b}{b^2-2b+1} - \frac{4}{b^2-2b+1} & = \frac{b^2+3b}{\color{blue}{b^2-2b+1}} - \frac{4}{\color{blue}{b^2-2b+1}} = \\[1ex] &=\frac{ b^2+3b - 4 }{ \color{blue}{ b^2-2b+1 }} \end{aligned} $$

Simplify $ \dfrac{b^2+3b-4}{b^2-2b+1} $ to $ \dfrac{b+4}{b-1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b-1}$.

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