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प्रश्न
Evaluate the following integrals : `int (2x + 3)/(2x^2 + 3x - 1).dx`
उत्तर
Let I = `int (2x + 3)/(2x^2 + 3x - 1).dx`
Let 2x + 3 = `"A"[d/dx(2x^2 + 3x - 1)] + "B"`
= A(4x + 3) + B
∴ 2x + 3 = 4Ax + (3A + B)
Comapring the coefficientof x and constant on both sides, we get
4A = 2 and 3A + B = 3
∴ A = `(1)/(2) and 3(1/2) + "B"` = 3
∴ B = `(3)/(2)`
∴ 2x + 3 = `(1)/(2)(4x + 3) + (3)/(2)`
∴ I = `int (1/2(4x + 3) + (3)/(2))/(2x^2 + 3x - 1).dx`
= `(1)/(2) int (4x + 3)/(2x^2 + 3x - 1).dx + (3)/(2) int (1)/(2x^2 + 3x - 1).dx`
= `(1)/(2)"I"_1 + (3)/(2)"I"_2`
I1 is of the type `int (f'(x))/f(x)dx = log|f(x)| + c`
∴ I1 = log |2x2 + 3x – 1| + c1
I2 = `int (1)/(2x^2 + 3x - 1).dx`
= `(1)/(2) int (1)/(x^2 + 3/2x - 1/2).dx`
= `(1)/(2) int (1)/((x^2 + 3/2x + 9/16) - 9/16 - 1/2).dx`
= `(1)/(2) int (1)/((x + 3/4)^2 - (sqrt(17)/4)^2).dx`
= `(1)/(2) xx (1)/(2 xx sqrt(17)/(4))log|(x + 3/4 - sqrt(17)/4)/(x + 3/4 + sqrt(17)/4)| + c_2`
= `(1)/sqrt(17)log|(4x + 3 - sqrt(17))/(4x + 3 + sqrt(17))| + c_2`
∴ I = `(1)/(2)log|2x^2 + 3x - 1| + (3)/(2sqrt(17))log|(4x + 3 - sqrt(17))/(4x + 3 + sqrt(17))| + c`, where c = c + c2.
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