Question

Integrate the following with respect to x:\

`logx/(1 + log)^2`

Answer 1

I = `int logx/(1 + log x)^2  "d"x`

Put t = ` log x`

⇒ x = e^t

dx = e^t dt

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

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

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

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

f(t) = `1/(1 + "t")`

f'(t) = `- 1/(1 + "t")^2`

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

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

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