Question

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

Options

• `tan^-1 x - 1/(2x(x^2 + 1)) + "c"`

• `1/2 tan^-1 x + x/(2(x^2 + 1)) + "c"`

• `tan^-1 x + 1/(x^2 + 1) + "c"`

• `tan^-1 x + 1/(2(x^2 + 1)) + "c"`

Answer 1

`int 1/(x^2 + 1)^2`dx = `underline(1/2 tan^-1 x + x/(2(x^2 + 1)) + "c")`.

Explanation:

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

`int 1 "dx" - int ("d"/"dx" (1/(x^2 + 1)) int 1 "dx") "dx"`    ...(Integration by parts)

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

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

`= x/(x^2 + 1) + 2 int x/(x^2 + 1) "dx" - 2 int "dx"/(x^2 + 1)^2`   ....(i)

And also, we know that

`int "dx"/(x^2 + 1)  = tan^-1 (x)`   ...(ii)

From Eqs. (i) and (ii), we get

`tan^-1 x = x/(x^2 + 1) + 2tan^-1 x - 2 int "dx"/(x^2 + 1)^2`

`=> 2 int "dx"/(x^2 + 1)^2 = tan^-1 x + x/(x^2 + 1)`

`=> int "dx"/(x^2 + 1)^2 = 1/2 tan^-1 x + x/(2((x^2 + 1)))`+ c