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प्रश्न
Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`
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उत्तर
Let I = `int (3x + 1)/sqrt(-2x^2 + x + 3).dx`
Let 3x + 1 = `"A"[d/dx(-2x^2 + x + 3)] + "B"`
= A(2 – 2x) + B
∴ 3x + 1 = 2Ax + (2A + B)
Comparing the coefficient of x and constant on both the sides, we get
– 2A = 7 and 2A + B = 3
∴ A = `(-7)/(2) and 2(-7/2) + "B" ` = 3
∴ B = 10
∴ 7x + 3 = `(-7)/(2)(2 - 2x) + 10`
∴ I = `int ((-7)/(2)(2 - 2x) + 10)/sqrt(3 + 2x - x^2).dx`
= `(-7)/(2) int ((2 - 2x))/sqrt(3 + 2x - x^2).dx + 10 int(1)/sqrt(3 + 2x - x^2)x`
= `(-7)/(2)"I"_1 + 10"I"_2`
In I1, put 3 + 2x – x2 = t
∴ (2 – 2x)dx = dt
∴ I1 = `int (1)/sqrt(t)dt`
= `int t^(-1/2) dt`
= `t^(1/2)/(1/2) + c_1`
= `2sqrt(3 + 2x - x^2) + c_1`
I2 = `int (1)/sqrt(3 - (x^2 - 2x + 1) + 1).dx`
= `int (1)/sqrt((2)^2 - (x - 1)^2).dx`
= `sin^-1((x - 1)/2) + c_2`
`-(3)/(2) sqrt(-2x^2 + x + 3) + (7)/(4sqrt(2)) sin^-1((4x - 1)/5) + c`.
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