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

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+3 }$ and the second by $\color{blue}{ x+1 }$.

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