Solve g(x)=1/x-1+5/x+1-2x/x^2-1

$$g ( x ) = \frac { 1 } { x - 1 } + \frac { 5 } { x + 1 } - \frac { 2 x } { x ^ { 2 } - 1 }$$

Evaluate

$$\frac{4}{x+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-1 and x+1 is \left(x-1\right)\left(x+1\right). Multiply \frac{1}{x-1} times \frac{x+1}{x+1}. Multiply \frac{5}{x+1} times \frac{x-1}{x-1}.

$$\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{x^{2}-1}$$

Since \frac{x+1}{\left(x-1\right)\left(x+1\right)} and \frac{5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{x+1+5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{x^{2}-1}$$

Do the multiplications in x+1+5\left(x-1\right).

$$\frac{x+1+5x-5}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{x^{2}-1}$$

Combine like terms in x+1+5x-5.

$$\frac{6x-4}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{x^{2}-1}$$

Factor x^{2}-1.

$$\frac{6x-4}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}$$

Since \frac{6x-4}{\left(x-1\right)\left(x+1\right)} and \frac{2x}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6x-4-2x}{\left(x-1\right)\left(x+1\right)}$$

Combine like terms in 6x-4-2x.

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

Factor the expressions that are not already factored in \frac{4x-4}{\left(x-1\right)\left(x+1\right)}.

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

Cancel out x-1 in both numerator and denominator.

$$\frac{4}{x+1}$$

Differentiate w.r.t. x

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