Advertisements
Advertisements
Question
Integrate the following with respect to x
`1/(x log x)`
Solution
Let log x = z
Then `1/x "d"x` = dz
So `int 1/(x log x) "d"x = int ("d"z)/z = log z + "c"`
= log |log x| + c
APPEARS IN
RELATED QUESTIONS
Integrate the following with respect to x.
(3 + x)(2 – 5x)
Integrate the following with respect to x.
`1/(sqrt(x + 1) + sqrt(x - 1))`
Integrate the following with respect to x.
`(sqrt(2x) - 1/sqrt(2x))^2`
Integrate the following with respect to x.
`(x^3 + 3x^2 - 7x + 11)/(x + 5)`
Integrate the following with respect to x.
`1/(x^2 + 3x + 2)`
Integrate the following with respect to x.
`1/(x + sqrt(x^2 - 1)`
Choose the correct alternative:
`int logx/x "d"x, x > 0` is
Evaluate the following integral:
`int 1/(sqrt(x + 2) - sqrt(x + 3)) "d"x`
Evaluate the following integral:
`int ("d"x)/("e"^x + 6 + 5"e"^-x)`
Evaluate the following integral:
`int_0^1 sqrt(x(x - 1)) "d"x`
