Question

Evaluate the following integrals:

`int (7x + 3)/sqrt(3 + 2x - x^2).dx`

Answer 1

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