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Evaluate the following integrals : ∫3x+42x2+2x+1.dx - Mathematics and Statistics

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प्रश्न

Evaluate the following integrals :  `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`

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उत्तर

Let I = `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`

Let 3x + 4 = `"A"[d/dx (2x^2 + 2x + 1)\ + "B"`   ...(i)

3x + 4 = A(4x + 2) + B
∴ 3x + 4 = (4A)x + (2A + B)
Consider,
4A = 3 and 2A + B = 4

∴ A = `(3)/(4) and 2(3/4) + "B"` = 4

∴ B = `4- 3/2`

∴ B = `8 - 3/2`

∴ B = `(5)/(2)`

From (i),

(3x + 4) = `3/4 d/dx (2x^2 + 2x + 1) + 5/2`   ...(ii)

The required integral is, 

I = `int ((3/4.d/dx (2x^2 + 2x + 1) + 5/2)/(sqrt(2x^2 + 2x + 1))dx`

I = `3/4 int (d/dx (2x^2 + 2x + 1))/(sqrt(2x^2 + 2x + 1)) dx + 5/2 int 1/ (sqrt(2x^2 + 2x + 1))dx`

I = `3/4 . 2 . sqrt(2x^2 + 2x + 1) + 5/2 . 1/sqrt2 int 1/sqrt(x^2 + x + 1/2)dx + c_1`  ...`int(f'(x))/sqrtf(x)dx = 2 sqrtf(x) + c`

I = `3/2 sqrt(2x^2 + 2x + 1) + 5/(2sqrt2) int 1/sqrt((x^2 + x + 1/4) + 1/2 - 1/4)dx + c_1` 

I = `3/2 sqrt(2x^2 + 2x + 1) + 5/(2sqrt2) int 1/ sqrt((x + 1/2)^2 + (1/2)^2)dx + c_1`

I = `3/2 sqrt(2x^2 + 2x + 1) + 5/(2sqrt2) log |(x + 1/2) + sqrt((x + 1/2)^2 + (1/2)^2)| + c_1 + c_2`

I = `3/2 sqrt(2x^2 + 2x + 1) + 5/(2sqrt2) log |(x + 1/2) + sqrt(x^2 + x + 1/2)| + c`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.2 (C) [पृष्ठ १२८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.2 (C) | Q 1.4 | पृष्ठ १२८

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