The value of `int_0^oo (dx)/(1+x^4)` is

Answer

B, C

Explanation

`I=int_(0)^(oo) (dx)/(1+x^(4))` ……………1
`=int_(0)^(oo) (x^(2)+1-x^(2))/(1+x^(4))dx`
`=int_(0)^(oo)(x^(2))/(1+x^(4))dx+int_(0)^(oo)(1-x^(2))/(1+x^(4))dx=I_(1)+I_(2)`
`I_(2)=int_(0)^(oo) (1/(x^(2))-1)/(1/(x^(2))+x^(2))dx`
Put `x+1/x=y`
`:.I_(2)=int_(oo)^(oo) (-1)/(y^(2)-2)dy=0`
`:.I=int_(0)^(oo) (dx)/(1+x^(4))=int_(0)^(oo)(x^(2)dx)/(1+x^(4))`
Adding 1 and 2 we get
`2I=int_(0)^(oo) (1+x^(2)dx)/(1+x^(4))=int_(0)^(oo) (1/(x^(2))+1)/(1/(x^(2))+x^(2))dx` (Putting `x=1/x=`)
`=int_(-oo)^(oo) (dy)/(y^(2)+2)`
`=[1/(sqrt(2))"tan"^(-1)y/(sqrt(2))]_(-oo)^(oo) =(pi)/(sqrt(2))`
or `(pi)/(2sqrt(2))`