Let `u=int_0^oo (dx)/(x^4+7x^2+1` and `v=int_0^oo (x^2dx)/(x^4+7x^2+1)` then

Options

A. `v>u`
B. `6v=pi`
C. `3u+2v=(5pi)/6`
D. `u+v=pi/3`

Answer

B

Explanation

`u=int_(0)^(oo) (dx)/(x^(4)+7x^(2)+1)` and `v=int_(0)^(oo) (x^(2)dx)/(x^(4)+7x^(2)+1)`
`:. u+v=int_(0)^(oo) (1+x^(2))/(x^(4)+7x^(2)+1)dx`
`=int_(0)^(oo) (1/(x^(2))+1)/((x-1/x)^(2)+9)dx`
`=1/3["tan"^(-1)((x-1/x)/3)]_(0)^(oo)`
`=1/3[pi//2+pi//2]=pi//3`
`:.u+v=pi//3`
Now `u-v=int_(0)^(oo) (1-x^(2))/(x^(4)+7x^(2)+1)dx`
Let `x=1/t` or `x=-(dt)/(t^(2))`
`:.u-v=int_(oo)^(0)(1-1/(t^(2)))/(1/(t^(4))+7/(t^(2))+1)(1-1/(t^(2)))dt`
`=-int_(0)^(oo)(1-t^(2))/(t^(4)+7t^(2)+1) dt`
`=-(u-v)`
`:. u-v=0`
From 1 and 2, we get `u=v=pi//6`