Advertisements
Advertisements
Question
Evaluate the following integral:
`int log (x - sqrt(x^2 - 1)) "d"x`
Solution
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"`
APPEARS IN
RELATED QUESTIONS
Integrate the following with respect to x.
`(sqrt(2x) - 1/sqrt(2x))^2`
Integrate the following with respect to x.
`(x^4 - x^2 + 2)/(x - 1)`
Integrate the following with respect to x.
`sqrt(1 - sin 2x)`
Integrate the following with respect to x.
`1/(x^2 - x - 2)`
Integrate the following with respect to x.
`1/sqrt(x^2 + 6x + 13)`
Choose the correct alternative:
`int "e"^x/("e"^x + 1) "d"x` is
Choose the correct alternative:
`int[9/(x - 3) - 1/(x + 1)] "d"x` is
Evaluate the following integral:
`int ("d"x)/("e"^x + 6 + 5"e"^-x)`
Evaluate the following integral:
`int sqrt(2x^2 - 3) "d"x`
Evaluate the following integral:
`int_(-1)^1 x^2 "e"^(-2x) "d"x`
