Question

Integrate the following with respect to x:

`(x + 2)/sqrt(x^2 - 1)`

Answer 1

Let x + 2 = `"A" "d"/("d"x) (x^2 - 1) + "B"`

x + 2 = A(2x) + B

2A = 1

⇒ A = `1/2`

B = 2

x + 2 = `1/2 (2x) + 2`

`int (x + 2)/sqrt(x^2 - 1)  "d"x = int (1/2 (2x) + 2)/sqrt(x^2 - 1)  "d"x`

= `1/2 int (2x)/sqrt(x^2 - 1)  "d"x + 2 int ("d"x)/sqrt(x^2 - 1)`

Put `x^2 - 1` = t

2x dx = dt

= `1/2 int "dt"/"t" + 2 int ("d"x)/sqrt(x^2 - 1^2)`

= `1/2 "t"^(- 1/2)  "dt" + 2 log |x + sqrt(x^2 - 1^2)| + "c"`

= `1/2 ("t"^(- 1/2 + 1))/(- 1/2 + 1) + 2 log |x + sqrt(x^2 - 1)| + "c"`

= `1/2 ("t"^(1/2))/(1/2) + 2 log |x + sqrt(x^2 - 1)| + "c"`

`int (x + 2)/sqrt(x^2 - 1)  "d"x = sqrt(x^2 - 1) + 2 log |x + sqrt(x^2 - 1)| + "c"`