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Evaluate: ∫x2+x-1x2+x-6) dx

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प्रश्न

Evaluate: `int ("x"^2 + "x" - 1)/("x"^2 + "x" - 6)` dx

योग
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उत्तर

Let I = `int ("x"^2 + "x" − 1)/("x"^2 + "x" − 6)` dx

`= int(("x"^2 + "x" − 6) + 5)/("x"^2 + "x" − 6)` dx

`= int [("x"^2 + "x" − 6)/("x"^2 + "x" − 6) + 5/("x"^2 + "x" − 6)]` dx

`= int [1 + 5/("x"^2 + "x" − 6)]` dx

`int [1 + 5/(("x + 3")("x − 2"))]` dx

Let `5/(("x + 3")("x − 2")) = "A"/"x + 3" + "B"/"x − 2"`

∴ 5 = A(x − 2) + B(x + 3)   ....(i)

Putting x = 2 in (i), we get

5 = A (0) + B (5)

∴ 5 = 5B

∴ B = 1

Putting x = − 3 in (i), we get

5 = A(− 5) + B (0)

∴ 5 = − 5A

∴ A = − 1

∴ `5/(("x + 3")("x - 2")) = (-1)/"x + 3" + 1/"x − 2"`

∴ I = `int [1 + (-1)/"x + 3" + 1/"x − 2"]` dx

`= int "dx" - int 1/"x + 3" "dx" + int1/"x − 2"` dx

∴ I = x − log |x + 3| + log |x − 2| + c

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अध्याय 5: Integration - EXERCISE 5.6 [पृष्ठ १३५]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 5 Integration
EXERCISE 5.6 | Q 3) | पृष्ठ १३५

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