Question

Integrate the following functions with respect to x :

`(3x - 9)/((x - 1)(x + 2)(x^2 + 1))`

Answer 1

`(3x - 9)/((x - 1)(x + 2)(x^2 + 1)) = "A"/(x - 1) + "B"/(x + 2) + ("C"x + "D")/(x + 2)^2`

3x – 9 = A(x + 2)(x² + 1) + B(x – 1) (x² + 1) + (Cx + D) (x – 1)(x + 2)

3x – 9 = A(x + 2)(x² + 1) + B(x – 1) + (x² + 1) + Cx(x – 1)(x + 2) + D(x – 1)(x + 2)

Put x = – 2

3 × – 2 – 9 = A(– 2 + 2)((2)² + 1) + B(– 2 – 1)((– 2)² + 1) + C(– 2)(– 2 – 1)(– 2 + 2) + D(– 2 – 1)(– 2 + 2)

– 6 – 9 = A × 0 + B × (– 3)(4 + 1) + C × 0 + D × 0

– 15 = B × – 3 × 5

– 15 = – 15B

⇒ B = 1

Put x = 1

3 × 1 – 9 = A(1 + 2)(12 + 1) + B(1 – 1)(12 + 1) + C × 1 (1 – 1)(1 + 2) + D(1 – 1)(1 + 2)

3 – 9 = A × 3 × 2 + B × 0 + C × 0 + D × 0

– 6 = 6A

⇒ A = – 1

Put x = 0

3 × 0 – 9 = A(0 + 2)(0² + 1) + B(0 – 1)(0² + 1) + C × 0 (0 – 1)(0 + 2) + D(0 – 1)(0 + 2)

– 9 = 2A – B + 0 – 2D

– 9 = 2A – B – 2D

– 9 = – 2 × – 1 – 1 – 2D

– 9 = – 2 – 1 – 2D

9 = 3 + 2D

⇒ 2D = 9 – 3

⇒ 2D = 6

⇒ D = 3

Put x = – 1

3 × – 1 – 9 = A(– 1 + 2)((1)² + 1) + B(– 1 – 1)((- 1)² + 1)) + C × – 1 × (– 1 – 1)(– 1 + 2) + D(– 1 – 1)(– 1 + 2)
– 3 – 9

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

– 12 = 2A – 4B + 2C – 2D

– 12 = 2 × – 1 – 4 × 1 + 2C – 2 × 3

– 12 = – 2 – 4 + 2C – 6

– 12 = – 12 + 2C

⇒ C = 0

`(3x - 9)/((x - 1)(x + 2)(x^2 + 1)) = (- 1)/(x - 1) + 1/(x + 2) + (0 * x + 3)/(1 + x^2)`

`int (3x - 9)/((x - 1)(x + 2)(x^2 + 1)) "d"x = int ((-1)/(x - 1) + 1/(x + 2) + 3/(1 + x^2)) "d"x`

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

= `- log |x - 1| + log |x + 2| + 3 tan^-1 x + "c"`

= `log |x + 2| + log|x - 1| + 3 tan^-1 x + "c"`

= `log |(x + 2)/(x - 1)| + 3 tan^-1 x + "c"`