Question

Evaluate the following integral:

`int log (x - sqrt(x^2 - 1)) "d"x`

Answer 1

Let u = `log(x - sqrt(x^2 - 1))`

du = `(1 - (2x)/(2sqrt(x^2 - 1)))/(x - sqrt(x^2 - 1))`

du = `(2sqrt(x^2 - 1) - 2x)/(2sqrt(x^2 - 1) (x - sqrt(x^2 - 1))`

du = `(-2(x - sqrt(x^2 - 1)))/(2sqrt(x^2 - 1)(x - sqrt(x^2 - 1))`

du = `(-1)/sqrt(x^2 - 1)`

So integral becomes

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

= `xlog(x - sqrt(x^2 - 1)) + 1/2 (x^2 - 1)^(1/2)/(1/2) + "c"`

= `xlog(x - sqrt(x^2 - 1)) + sqrt(x^2 - 1) + "c"`