how can be calculated the definite integral of ln((|x|+1)/(x^2+1)) if limits of integration are -1,1? |x| is absolute value

`f(x)= ln
(x+1)/(x^2+1)`

`==> f(x)= ln(x+1) -
ln(x^2+1)`

`==> int f(x)= int ln(x+1) - int
ln(x^2+1)`

`u= x+1 ==> du=
dx`

`int ln(x+1) dx = int ln u du = u*lnu - u + c
`

`==> int ln(x+1) dx = (x+1)*ln(x+1) - (x+1)
......(1)`

`int ln(x^2+1) dx
`

`u= ln(x^2+1) dx ==> du= 2x/(x^2+1)
dx`

`dv = dx ==> v =
x`

`int ln(x^2+1) dx = u*v- int
v*du`

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

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

`int ln(x^2+1) dx = xln(x^2+1) - 2 (x-arctan(x)
`

`int ln(x^2+1) dx = xln(x^2+1) - 2x - 2arctan(x) +
c.............(2)`

`int ln(x+1)- int ln(x^2+1) =
(x+1)ln(x+1) - (x+1) - xln(x^2+1) + 2x + 2arctan(x) +
C`

`int ln((x+1)/(x^2+1))= (x+1)ln(x+1) -xln(x^2+1) +
2arctan(x) +x + 1 + C`

`int_-1^1f(x)dx = int_-1^1
ln((x+1)/(x^2+1)) dx = (2ln2-2ln2 +2arctan(1) + 1+1 - [ 0+ln(2) + 2arctan(-1) -1
+1}`

` = 2arctan(1) +2- ln2 +
2arctan(-1)`

`=2*pi/4 + 2 - ln2 +
2*-pi/4`

`= 2 -
ln2`

``

``