How do you express as a partial fraction $x/((x+1)(x^2 -1))$ ?

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Answer 1 · 2015-08-22T02:58:48

$(-1/4)/(x+1) + (1/2)/(x+1)^2 + (1/4)/(x-1) = 1/4[(-1)/(x+1) + (2)/(x+1)^2 + 1/(x-1)]$

First we need to finish factoring the denominator into factors that are irreducible using Real coefficients:

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

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

So we need:

$A/(x+1) + B/(x+1)^2 + C/(x-1) = x/((x+1)^2(x-1))$

Clear the denominator (multiply by $((x+1)^2(x-1))$ on both sides), to get:

$A(x^2-1)+B(x-1)+C(x+1)^2 = x$

$Ax^2-A + Bx -B +Cx^2+2Cx+C = x$

$Ax^2+Cx^2 +Bx +2Cx -A-B+C = 0x^2 +1x+0$

So we need to solve:

$A+C = 0$
$B+2C = 1$
$-A-B+C=0$

From the first equation, we get: $A = -C$ and we can substitue in the third equation to get

Eq 2: $B+2C = 1$
and: $-B+2C = 0$

Adding gets us $C = 1/4$ , so $A = -1/4$ and subtracting gets us $B = 1/2$

$A/(x+1) + B/(x+1)^2 + C/(x-1) = (-1/4)/(x+1) + (1/2)/(x+1)^2 + (1/4)/(x-1)$

$= 1/4[(-1)/(x+1) + (2)/(x+1)^2 + 1/(x-1)]$

How do you express as a partial fraction x/((x+1)(x^2 -1))x/((x+1)(x^2 -1))?