Solve x+1/x^2-2x+x+1/x^2+2x-2x/x^2-4

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

Evaluate

$$\frac{2}{x^{2}-4}$$

Short Solution Steps

Factor x^{2}-2x. Factor x^{2}+2x.

$$\frac{x+1}{x\left(x-2\right)}+\frac{x+1}{x\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x\left(x-2\right) and x\left(x+2\right) is x\left(x-2\right)\left(x+2\right). Multiply \frac{x+1}{x\left(x-2\right)} times \frac{x+2}{x+2}. Multiply \frac{x+1}{x\left(x+2\right)} times \frac{x-2}{x-2}.

$$\frac{\left(x+1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Since \frac{\left(x+1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)} and \frac{\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+1\right)\left(x+2\right)+\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Do the multiplications in \left(x+1\right)\left(x+2\right)+\left(x+1\right)\left(x-2\right).

$$\frac{x^{2}+2x+x+2+x^{2}-2x+x-2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Combine like terms in x^{2}+2x+x+2+x^{2}-2x+x-2.

$$\frac{2x^{2}+2x}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Factor the expressions that are not already factored in \frac{2x^{2}+2x}{x\left(x-2\right)\left(x+2\right)}.

$$\frac{2x\left(x+1\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Cancel out x in both numerator and denominator.

$$\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Factor x^{2}-4.

$$\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}$$

Since \frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)} and \frac{2x}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(x+1\right)-2x}{\left(x-2\right)\left(x+2\right)}$$

Do the multiplications in 2\left(x+1\right)-2x.

$$\frac{2x+2-2x}{\left(x-2\right)\left(x+2\right)}$$

Combine like terms in 2x+2-2x.

$$\frac{2}{\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-2\right)\left(x+2\right).

$$\frac{2}{x^{2}-4}$$

Expand

$$\frac{2}{x^{2}-4}$$

Short Solution Steps

Factor x^{2}-2x. Factor x^{2}+2x.

$$\frac{x+1}{x\left(x-2\right)}+\frac{x+1}{x\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x\left(x-2\right) and x\left(x+2\right) is x\left(x-2\right)\left(x+2\right). Multiply \frac{x+1}{x\left(x-2\right)} times \frac{x+2}{x+2}. Multiply \frac{x+1}{x\left(x+2\right)} times \frac{x-2}{x-2}.

$$\frac{\left(x+1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Since \frac{\left(x+1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)} and \frac{\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+1\right)\left(x+2\right)+\left(x+1\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Do the multiplications in \left(x+1\right)\left(x+2\right)+\left(x+1\right)\left(x-2\right).

$$\frac{x^{2}+2x+x+2+x^{2}-2x+x-2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Combine like terms in x^{2}+2x+x+2+x^{2}-2x+x-2.

$$\frac{2x^{2}+2x}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Factor the expressions that are not already factored in \frac{2x^{2}+2x}{x\left(x-2\right)\left(x+2\right)}.

$$\frac{2x\left(x+1\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Cancel out x in both numerator and denominator.

$$\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{x^{2}-4}$$

Factor x^{2}-4.

$$\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}$$

Since \frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)} and \frac{2x}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(x+1\right)-2x}{\left(x-2\right)\left(x+2\right)}$$

Do the multiplications in 2\left(x+1\right)-2x.

$$\frac{2x+2-2x}{\left(x-2\right)\left(x+2\right)}$$

Combine like terms in 2x+2-2x.

$$\frac{2}{\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-2\right)\left(x+2\right).

$$\frac{2}{x^{2}-4}$$