Question

`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?

पर्याय

• `x/(1 + (log x)^2)`+ c

• `x/(1 + (log x))`+ c

• `x^2/(1 + (log x)^2)`+ c

• `1/(1 + (log x)^2)`+ c

Answer 1

`x/(1 + (log x)^2)`+ c

Explanation:

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

Put, logx = t ⇒ x = e^f ⇒ dx = e^t dt

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

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

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

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

`therefore "I" = "e"^"t"/(1 + "t"^2)` + c   ....[∵ ∫e^x f(x) + f'(x) dx = e^x f(x) + c]

`=> "I" = x/(1 + (log x)^2)` + c