Advertisements
Advertisements
प्रश्न
Integrate the following with respect to x
`1/(x log x)`
उत्तर
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
संबंधित प्रश्न
Integrate the following with respect to x.
`(x^3 + 3x^2 - 7x + 11)/(x + 5)`
Integrate the following with respect to x.
`"e"^(xlog"a") + "e"^("a"log"a") - "e"^("n"logx)`
Integrate the following with respect to x.
`"e"^(2x)/("e"^(2x) - 2)`
Integrate the following with respect to x.
`1/(x^2(x^2 + 1))`
Choose the correct alternative:
`int 1/x^3 "d"x` is
Choose the correct alternative:
`int (sin2x)/(2sinx) "d"x` is
Choose the correct alternative:
`int (2x + 3)/sqrt(x^2 + 3x + 2) "d"x` is
Choose the correct alternative:
`int_0^1 sqrt(x^4 (1 - x)^2) "d"x` is
Evaluate the following integral:
`int 1/(sqrt(x + 2) - sqrt(x + 3)) "d"x`
Integrate the following with respect to x.
`sqrtx (x^3 - 2x + 3)`
