Question

Resolve the following rational expressions into partial fractions

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

Answer 1

`x/((x^2 + 1)(x - 1)(x + 2)) = ("A"x + "B")/(x^2 + 1) + "C"/(x - 1) + "D"/(x + 2)`

`x/((x^2 + 1)(x - 1)(x + 2)) = (("A"x + "B")(x - 1)(x + 2) + "C"(x^2 + 1)(x + 2) + "D"(x^2 + 1)(x - 1))/((x^2 + 1)(x - 1)(x + 2))`

x = A(x + 1)(x + 2) + B(x – 1)(x + 2) + C(x² + 1)(x + 2) + D(x² + 1)(x – 1)  ......(1)

Put x = 1 in equation (1)

1 = A(1)(1 – 1)(1 + 2) + B(1 – 1)(1 + 2) + C(1² + 1)(1 + 2) + D(1² + 1)(1 – 1)

1 = A × 0 + B × 0 + C(2)(3) + D × 0

1 = 6C

⇒ C = `1/6`

Put x = – 2 in equation (1)

– 2 = A(– 2)(– 2 – 1)(– 2 + 2) + B(– 2 – 1)(– 2 + 2) + C ((– 2)² + 1)(– 2 + 2) + D((– 2)² + 1)(– 2 – 1)

– 2 = A × 0 + B × 0 + C × 0 + D(4 + 1)(– 3)

– 2 = D(5)(– 3)

⇒ – 2 = – 15 D

⇒ D = `2/15`

Put x = 0 in equation (1)

0 = A(0)(0 – 1)(0 + 2) + B(0 – 1)(0 + 2)+ C(0² + 1)(0 + 2) + D(0² + 1)(0 – 1)

0 = 0 + B(– 2) + C(2) + D(– 1)

0 = – 2B + 2C – D

0 = `- 2"B" + 2 xx 1/6 - 2/15`

⇒ 2B = `1/3 - 2/15`

⇒ 2B = `(5 - 2)/15`

⇒ 2B = `3/15`

⇒ 2B = `1/5`

⇒ B = `1/10`
In equation (1), equate the coefficient of x³ on both sides

0 = A + C + D

0 = `"A" + 1/6 + 2/15`

⇒ A = `- 1/6 - 2/15`

= `(- 5 - 4)/30`

= `- 9/30`

⇒ A = `- 3/10`

∴ The required partial fractions is

`x/((x^2 + 1)(x - 1)(x + 2)) = (- 3/10 x + 1/10)/(x^2 + 1) + (1/6)/(x - 1) + (2/15)/(x + 2)`

`x/((x^2 + 1)(x - 1)(x + 2)) = (1 - 3x)/(10(x^2 + 1)) + 1/(6(x - 1)) + 2/(15(x + 2))`