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

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

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x+1}$.

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

Divide $ \dfrac{-x}{x+1} $ by $ x $ to get $ \dfrac{ -x }{ x^2+x } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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