मराठी

∫x2-8x+7dx is equal to ______.

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प्रश्न

`int sqrt(x^2 - 8x + 7) dx` is equal to ______.

पर्याय

  • `1/2  (x - 4) sqrt(x^2 - 8x + 7) + 9 log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C`

  • `1/2  (x + 4) sqrt(x^2 - 8x + 7) + 9 log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C`

  • `1/2  (x - 4) sqrt(x^2 - 8x + 7) + 3sqrt2  log abs (x - 4 + sqrt((x^2 - 8x + 7))) + C`

  • `1/2  (x - 4) sqrt(x^2 - 8x + 7) - 9/2  log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C`

MCQ
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उत्तर

`int sqrt(x^2 - 8x + 7) dx` is equal to `underline(1/2  (x - 4) sqrt(x^2 - 8x + 7) - 9/2  log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C)`.

Explanation:

`int sqrt(x^2 - 8x + 7)  dx`

`= int sqrt(x^2 - 8x + 16 + 7 - 16)  dx`

`= int sqrt((x - 4)^2 - 9)  dx`

`= int sqrt(x^2 - a^2)  dx = x/2 sqrt (x^2 - a^2) - a^2/2  log abs (x + sqrt(x^2 - a^2)) + C`

On substituting x - 4 and a2 = 9 in place of x,

`therefore int sqrt((x - 4)^2 - 9)  dx = ((x - 4))/2 sqrt(x^2 - 8x + 7) - 9/2  log abs ((x - 4) + sqrt (x^2 - 8x + 7)) + C`

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पाठ 7: Integrals - Exercise 7.7 [पृष्ठ ३३०]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.7 | Q 11 | पृष्ठ ३३०

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