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

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{15}i}{4}\approx 0.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i+1}{4}\approx 0.25-0.968245837i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-1,x-1,x+1.

$$1+\left(x+1\right)x=\left(x-1\right)\left(1-x\right)$$

Use the distributive property to multiply x+1 by x.

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

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

$$1+x^{2}+x=2x-x^{2}-1$$

Subtract 2x from both sides.

$$1+x^{2}+x-2x=-x^{2}-1$$

Combine x and -2x to get -x.

$$1+x^{2}-x=-x^{2}-1$$

Add x^{2} to both sides.

$$1+x^{2}-x+x^{2}=-1$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$1+2x^{2}-x=-1$$

Add 1 to both sides.

$$1+2x^{2}-x+1=0$$

Add 1 and 1 to get 2.

$$2+2x^{2}-x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-16}}{2\times 2}$$

Add 1 to -16.

$$x=\frac{-\left(-1\right)±\sqrt{-15}}{2\times 2}$$

Take the square root of -15.

$$x=\frac{-\left(-1\right)±\sqrt{15}i}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{15}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{15}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{15}i}{4} when ± is plus. Add 1 to i\sqrt{15}.

$$x=\frac{1+\sqrt{15}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{15}i}{4} when ± is minus. Subtract i\sqrt{15} from 1.

$$x=\frac{-\sqrt{15}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+1}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-1,x-1,x+1.

$$1+\left(x+1\right)x=\left(x-1\right)\left(1-x\right)$$

Use the distributive property to multiply x+1 by x.

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

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

$$1+x^{2}+x=2x-x^{2}-1$$

Subtract 2x from both sides.

$$1+x^{2}+x-2x=-x^{2}-1$$

Combine x and -2x to get -x.

$$1+x^{2}-x=-x^{2}-1$$

Add x^{2} to both sides.

$$1+x^{2}-x+x^{2}=-1$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$1+2x^{2}-x=-1$$

Subtract 1 from both sides.

$$2x^{2}-x=-1-1$$

Subtract 1 from -1 to get -2.

$$2x^{2}-x=-2$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-1+\frac{1}{16}$$

Add -1 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{15}i}{4} x-\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+1}{4}$$