Question

Evaluate Δ`[1/((x + 1)(x + 2))]` by taking ‘1’ as the interval of differencing

Answer 1

Let `1/((x + 1)(x + 2)) = "A"/((x + 1)) + "B"/((x + 2))`

By Partial fraction method

A = `1/((x + 2)) [x = -1]`

= `1/(1 ++ 2)` = 1

A = 1

B = `1/((x + 1)) [x = -2]`

= `1/(-2 + 1)`

 = – 1

B = – 1

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

`[1/((x + 1) + (x + 2))] = Delta [1/((x + 1)) - 1/((x + 2))]`

= `[1/((x + 1 + 1)) - 1/((x + 1))] - [1/((x + 2 + 1)) - 1/((x + 2))]`

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

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

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

= `(2(x^2 + 4x + 3) - (x^2 + 5x + 6) - (x^2 + 3x + 2))/((x + 1)(x + 2)(x + 3))`

= `(2x^2 + 8x + 6 - x^2 - 5x - 6 - x^2 - 3x - 2)/((x + 1)(x + 2)(x + 3))`

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

`Delta[1/((x + 1)(x + 2))] = (-2)/((x + 1)(x + 2)(x + 3))`