Question

`int log x * [log ("e"x)]^-2` dx = ?

Options

• `x/(1 + log x) + "c"`

• x(1 - log x) + c

• x(1 + log x) + c

• `x/(1 - log x) + "c"`

Answer 1

`x/(1 + log x) + "c"`

Explanation:

Let I = `int log x * [log ("e"x)]^-2` dx

`= int (log x)/((log "e"x)^2)`dx

`= int (log x)/(log "e" + log x)^2`dx

`= int (log x)/(1 + log x)^2` dx

Put log x = t ⇒ x = e^t

⇒ dx = e^t dt

`= int "t"/(1 + "t")^2 "e"^"t" "dt"`

`= int "e"^"t" (1/(1 + "t") - 1/(1 + "t")^2)`dt

`= "e"^"t"/(1 + "t")` + c

`= x/(1 + log x) + "c"`