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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate: int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx

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प्रश्न

Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`

मूल्यांकन
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उत्तर

Consider, `(2x^2 - 3)/((x^2 - 5)(x^2 + 4))`

Let x2 = m

∴ `(2m - 3)/((m - 5)(m + 4))`  ...[proper rational function]

Now, `(2m - 3)/((m - 5)(m + 4)) = A/((m - 5)) + B/((m + 4))`

`(2m - 3)/((m - 5)(m + 4)) = (A(m + 4) + B(m - 5))/((m - 5)(m + 4))`

∴ 2m – 3 = A (m + 4) + B (m – 5)

at m = 5, 2(5) – 3 = A (9) + B (0)

7 = 9A `\implies` A = `7/9`

at m = –4, 2(–4) – 3 = A (0) + B (–9)

–11 = –9B `\implies` B = `11/9`

Thus, `(2m - 3)/((m - 5)(m + 4)) = ((7/9))/((m - 5)) + ((11/9))/((m + 4))` i.e. `(2x^2 - 3)/((x^2 - 5)(x^2 + 4)) = ((7/9))/((x^2 - 5)) + ((11/9))/((x^2 + 4))`

∴ I = `int [((7/9))/(x^2 - 5) + ((11/9))/(x^2 + 4)]dx`

= `7/9 . int 1/(x^2 - (sqrt(5))^2)dx + 11/9 . int 1/(x^2 + (2)^2)dx`

= `7/9 . 1/(2(sqrt(5))) . log [(x - sqrt(5))/(x + sqrt(5))] + 11/9 . 1/2 . tan^-1 (x/2) + c`

∴ I = `7/(18(sqrt(5))) . log [(x - sqrt(5))/(x + sqrt(5))] + 11/18 . tan^-1 (x/2) + c`

∴ `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx = 7/(18(sqrt5)) . log [(x - sqrt(5))/(x + sqrt(5))] + 11/18 . tan^-1 (x/2) + c`

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