Question

Evaluate the following integrals : `int sqrt((x - 7)/(x - 9)).dx`

Answer 1

Let I = `int sqrt((x - 7)/(x - 9)).dx`

= `int sqrt((x - 7)/(x - 9) xx (x - 7)/(x - 7)).dx`

= `int sqrt((x - 7)^2/(x^2 - 16x + 63)).dx`

= `int (x - 7)/sqrt(x^2 - 16x + 63).dx`

Let x – 7 = `"A"[d/dx(x^2 - 16x + 63)] + "B"`

= A(2x – 16) + B
= 2Ax + (B – 16A)
Comparing the coefficient of x and constant term on both sides, we get
2A = 1

∴ A = `(1)/(2)` and

B – 16A = – 7

∴ `"B" - 16(1/2)` = – 7
∴ B = 1
∴ x – 7 = `(1)/(2)(2x - 16) + 1`

∴ I = `int (1/2(2x - 16) + 1)/sqrt(x^2 - 16x + 63).dx`

 = `(1)/(2) int (2x - 16)/sqrt(x^2 - 16x + 63).dx + int (1)/sqrt(x^2 - 16x + 63).dx`

= `(1)/(2)"I"_1 + "I"_2`

In I₁, put x² – 16x + 63 = t

∴ (2x – 16)dx = dt

∴ I₁ = `(1)/(2) int (1)/sqrt(t)dt`

= `(1)/(2) int t^(-1/2)dt`

= `(1)/(2) t^(1/2)/((1/2)) + c_1`

= `sqrt(x^2 - 16x + 63) + c_1`

I₂ = `int (1)/sqrt(x^2 - 16x + 63).dx`

= `int (1)/sqrt((x - 8)^2 - 1^2).dx`

= `log|x - 8 + sqrt((x - 8)^2 - 1^2)| + c_2`

= `log|x  - 8 + sqrt(x^2 - 16x + 63)| + c_2`

∴ I = `sqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`,, where c = c₁ + c₂.