Question

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

Answer 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)]`

Explanation:

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)]`