How do you express $\frac{x}{(x - 1) (x + 1)}$ $x/((x-1)(x+1))$ in partial fractions?

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How do you express $\frac{x}{(x - 1) (x + 1)}$ $x/((x-1)(x+1))$ in partial fractions?

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Answer 1 · 2016-02-07T13:49:41

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

Explanation:

Let $\frac{x}{(x - 1) (x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$ $x/((x-1)(x+1))=(A)/(x-1)+(B)/(x+1)$

$\frac{x}{(x - 1) (x + 1)} = \frac{A}{x - 1} (\frac{x + 1}{x + 1}) + \frac{B}{x + 1} \cdot (\frac{x - 1}{x - 1})$ $x/((x-1)(x+1))=(A)/(x-1)((x+1)/(x+1))+(B)/(x+1)*((x-1)/(x-1))$

$\frac{x}{(x - 1) (x + 1)} = \frac{A x + A + B x - B}{(x - 1) (x + 1)}$ $x/((x-1)(x+1))=(Ax+A+Bx-B)/((x-1)(x+1))$

Equate the numerators

$x = A x + A + B x - B$ $x=Ax+A+Bx-B$

$x = A x + B x + A - B$ $x=Ax+Bx+A-B$

$1 \cdot x + 0 = (A + B) x + (A - B)$ $1*x+0=(A+B)x+(A-B)$

so that

$A + B = 1$ $A+B=1$ and $A - B = 0$ $A-B=0$

solving for A and B

$A = \frac{1}{2}$ $A=1/2$ and $B = \frac{1}{2}$ $B=1/2$

so therefore

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

have a nice day ! from the Philippines..

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How do you express $\frac{x}{(x - 1) (x + 1)}$ $x/((x-1)(x+1))$ in partial fractions?

1 Answer

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