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

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

Evaluate

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

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{1}{x+1} times \frac{x-1}{x-1}.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Expand \left(x-1\right)\left(x+1\right)\left(x^{2}+1\right).

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

Differentiate w.r.t. x

$$-\frac{8x\left(x^{4}+1\right)}{\left(x^{4}-1\right)^{2}}$$

Steps Using Derivative Rule for Quotient

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{1}{x+1} times \frac{x-1}{x-1}.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Use the distributive property to multiply x-1 by x+1 and combine like terms.

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

Consider \left(x^{2}-1\right)\left(x^{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4x^{2}}{x^{4}-1})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

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

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

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

Do the arithmetic.

$$\frac{\left(x^{4}-1\right)\times 8x^{1}-4x^{2}\times 4x^{3}}{\left(x^{4}-1\right)^{2}}$$

Expand using distributive property.

$$\frac{x^{4}\times 8x^{1}-8x^{1}-4x^{2}\times 4x^{3}}{\left(x^{4}-1\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{8x^{4+1}-8x^{1}-4\times 4x^{2+3}}{\left(x^{4}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{8x^{5}-8x^{1}-16x^{5}}{\left(x^{4}-1\right)^{2}}$$

Combine like terms.

$$\frac{\left(8-16\right)x^{5}-8x^{1}}{\left(x^{4}-1\right)^{2}}$$

Subtract 16 from 8.

$$\frac{-8x^{5}-8x^{1}}{\left(x^{4}-1\right)^{2}}$$

Factor out 8x.

$$\frac{8x\left(-x^{4}-x^{0}\right)}{\left(x^{4}-1\right)^{2}}$$

For any term t except 0, t^{0}=1.

$$\frac{8x\left(-x^{4}-1\right)}{\left(x^{4}-1\right)^{2}}$$