Correct Answer - B
Let `l = int_(0)^(oo) (x dx)/((1+x)(x^(2)+1))`
By Partial fraction,
`(x)/((1+x)(x^(2)+1))=(A)/((1+x))+(Bx+C)/((x^(2)+1))`
`rArr x=A(x^(2)+1)+(1+x)(Bx+C)`
`rArr x=A(x^(2)+1)+(Bx+Bx^(2)+C+Cx)`
`rArr x =(A+B)x^(2)+(B+C)x +(A+C)`
On comparing both sodes, we get
`A+B=0, B+C=1, A+C=0` ...(i)
On adding all these equations, we get
`A+B+C=(1)/(2)` ...(ii)
`therefore A=(1)/(2)-1=-(1)/(2),C=(1)/(2)` and `B=(1)/(2)`
Then, `l=int_(0)^(oo){(-1)/(2(1+x))+(1)/(2)((x+1))/((x^(2)+1))}dx`
`=-(1)/(2)int_(0)^(oo)(dx)/(1+x)+(1)/(2)int_(0)^(oo)(x)/(x^(2)+1)dx +(1)/(2)int_(0)^(oo)(dx)/(1+x^(2))`
`=-(1)/(2)[log(1+x)]_(0)^(oo)+(1)/(4)[log(x^(2)+1)]_(0)^(oo)+(1)/(2)xx(pi)/(2)`
`=-(1)/(2)lim_(x to oo)log (1+x)+(1)/(4) lim_(x to oo)log (1+x^(2))+(pi)/(4)`
`= lim_(x to oo)log [((1+x^(2))^(1//4))/((1+x)^(1//2))]+(pi)/(4)`
`= lim_(x to oo)log [(sqrt(x)((1)/(x^(2))+1)^(1//4))/(sqrt(x)((1)/(x)+1)^(1//2))]+(pi)/(4)`
`= log. ((0+1)^(1//4))/((0+1)^(1//2))+(pi)/(4)`
`= log (1) +(pi)/(4)=0+(pi)/(4)=(pi)/(4)`
