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Question
Evaluate the following integrals:
`int (7x + 3)/sqrt(3 + 2x - x^2).dx`
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Solution
Let I = `int (7x + 3)/sqrt(3 + 2x - x^2).dx`
Let 7x + 3 = `A[d/dx(3 + 2x - x^2)] + B`
= A(2 – 2x) + B
∴ 7x + 3 = -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`
∴ I = `-7sqrt(3 + 2x - x^2) + 10sin^-1((x - 1)/2) + c`, where c = c1 + c2
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