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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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