Question

Integrate the following functions with respect to x :

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

Answer 1

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

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

Put x= – 2

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

1 = A × 0 + B × 0 + C × – 3

C = `1/3`

Put x = 1

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

1 = A × 3² + B × 0 + C × 0

A = `1/9`

Put x = 0

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

1 = A × 4 – 2B – C

1 = `1/9 xx 4 - 2"B" + 1/3`

1 = `49 + 1/3 - 2"B"`

1 = `(4 + 3)/9 - 2"B"`

2B = `7/9 - 1`

2B = `(7 - 9)/9 = - 2/9`

B = `- 1/9`

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

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

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

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

=  `1/9 log |x - 1| - 1/9 log |x + 2| - 1/3 (x + 2)^(-2 + 1)/(- 2 + 1) + "c"`

= `1/9 log |x - 1| - 1/9 log |x + 2| - 1/3 (x + 2)^(-1)/(- 1) + "c"`

= `1/9 log |x - 1| - 1/9 log |x + 2| + 1/3 xx 1/(x + 2) + "c"`