# 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

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### 2 Answers

`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`

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Actually is ln of ratio ((|x|+1)/(x^2+1)).

why did you remove absolute value of x?

the answer is not i got in my book..