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