Question

If `intsqrt((x - 7)/(x - 9)) dx = Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, then A = ______

Options

• -1

• `1/2`

• `-1/2`

• 1

Answer 1

If `intsqrt((x - 7)/(x - 9)) dx = Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, then A = 1

Explanation:

Let I = `intsqrt((x - 7)/(x - 9)) dx`

= `intsqrt((x - 7)/((x - 9)(x - 7)))dx` 

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

= `1/2int(2x - 14)/(x^2 - 16x + 63)dx`

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

= `1/2int(2x - 16)/sqrt(x^2 - 16x + 63) + 2/2intdx/sqrt(x^2 - 16x + 64 - 1)`

= `1/2 xx 2sqrt(x^2 - 16x + 63) + int dx/sqrt((x - 8)^2 - 1)`

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

Comparing with `Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, we get

A = 1