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Answer

$$\frac{1}{x}-\frac{1}{x+1}\frac{x}{x^2+2x+1}+\frac{1}{x+1}=\frac{2x^3+4x^2+4x+1}{x^4+3x^3+3x^2+x}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{1}{x}-\frac{1}{x+1}\frac{x}{x^2+2x+1}+\frac{1}{x+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{x}-\frac{x}{x^3+3x^2+3x+1}+\frac{1}{x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^3+2x^2+3x+1}{x^4+3x^3+3x^2+x}+\frac{1}{x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2x^3+4x^2+4x+1}{x^4+3x^3+3x^2+x}\end{aligned} $$
①

Multiply $ \dfrac{1}{x+1} $ by $ \dfrac{x}{x^2+2x+1} $ to get $ \dfrac{x}{x^3+3x^2+3x+1} $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

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

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

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+3x^2+3x+1 }$ and the second by $\color{blue}{ x }$.

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

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

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x^3+2x^2+x}$.

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