The partial fractions of the expression `(2x - 1)/(3x^2+2x-1)` have to be determined.

`(2x - 1)/(3x^2+2x-1)`

First factor the denominator.

= `(2x - 1)/(3x^2 + 3x - x - 1)`

= `(2x - 1)/(3x(x + 1) - 1(x + 1))`

= `(2x - 1)/((3x - 1)(x + 1))`

This can...

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The partial fractions of the expression `(2x - 1)/(3x^2+2x-1)` have to be determined.

`(2x - 1)/(3x^2+2x-1)`

First factor the denominator.

= `(2x - 1)/(3x^2 + 3x - x - 1)`

= `(2x - 1)/(3x(x + 1) - 1(x + 1))`

= `(2x - 1)/((3x - 1)(x + 1))`

This can be written `A/(3x - 1) + B/(x +1)`

`A/(3x - 1) + B/(x +1) = (2x - 1)/((3x - 1)(x + 1))`

=> `A*(x + 1) + B*(3x - 1) = 2x - 1`

=> `Ax + A + 3Bx - B = 2x - 1`

=> `A + 3B = 2` and `A - B = -1`

`A - B = -1`

=> `A = B - 1`

Substitute in `A + 3B = 2`

=> `B - 1 + 3B = 2`

=> `4B = 3`

=> `B = 3/4`

`A = -1/4`

This gives: `3/(4*(x+1)) - 1/(4*(3x-1))`

**The partial fraction decompostion of `(2x - 1)/(3x^2+2x-1)` is **`3/(4*(x+1)) - 1/(4*(3x-1))`