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X + 2xxdxpxxxdxdxxx∫x + 22x2+6x+5dx=p∫4x+62x2+6x+5dx+12∫dx2x2+6x+5, then p = ?

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प्रश्न

`int ("x + 2")/(2"x"^2 + 6"x" + 5)"dx" = "p" int (4"x" + 6)/(2"x"^2 + 6"x" + 5) "dx" + 1/2 int "dx"/(2"x"^2 + 6"x" + 5)`, then p = ?

विकल्प

  • `1/3`

  • `1/2`

  • `1/4`

  • 2

MCQ
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उत्तर

`1/4`

Explanation:

Let x + 2 = p `"d"/"dx" (2"x"^2 + 6"x" + 5) + "q"`

= p(4x + 6) + q

∴ x + 2 = 4px + 6p + q

∴ 4p = 1 and 6p + q = 2

∴ p = `1/4`

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अध्याय 5: Integration - MISCELLANEOUS EXERCISE - 5 [पृष्ठ १३७]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 5 Integration
MISCELLANEOUS EXERCISE - 5 | Q I. 4) | पृष्ठ १३७

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