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∫ x2(x2+1)(x2-2)(x2+3) dx - Mathematics and Statistics

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प्रश्न

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

योग

उत्तर

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

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

= `"A"/(x^2 + 1) + "b"/(x^2 - 2) + "c"/(x^2 + 3)`

∴ x2 = A(x2 − 2)(x2 + 3) + B(x2 + 1)(x2 + 3) + C(x2 + 1)(x2 − 2)     ........(i)

Putting x2 = 2 in (i), we get

2 = B × 3 × 5

∴ B = `2/15`

Putting x2 = −3 in (i), we get

−3 = C × (– 2) × (– 5)

∴ C = `(-3)/10`

Putting x2 = −1 in (i), we get

−1 = A × (–3) × 2

∴ A = `1/6`

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

∴ I = `int[1/(6(x^2 + 1)) + 2/(15(x^2 - 2)) - 3/(10(x^2 + 3))]  "d"x`

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

= `1/6 int 1/(x^2 + 1)  "d"x + 2/15 int  1/(x^2 - (sqrt(2))^2)  "d"x - 3/10 int 1/(x^2 + (sqrt(3))^2)  "d"x`

= `1/6 tan^-1x + 2/15 xx 1/(2 xx sqrt(2)) log|(x - sqrt(2))/(x + sqrt(2))| - 3/10 xx 1/sqrt(3) tan^-1 (x/sqrt(3)) + "c"`

∴ I = `1/6 tan^-1x + 1/(15sqrt(2)) log|(x - sqrt(2))/(x + sqrt(2))| - sqrt(3)/10 tan^-1 (x/sqrt(3)) + "c"`

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अध्याय 2.3: Indefinite Integration - Long Answers III

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