If u=∫0∞dxx4+7x2+1 and v=∫0∞x2dxx4+7x2+1 then

u=∫0∞1x4+7x2+1dx Put x=1t,dx=−1t2=dtu=∫0∞11t4+7t2+1−dtt2=∫0∞t2dtt4+7t2+1=v⇒u=vSo, u+v=2u=∫0∞1+x2x4+7x2+1⇒2u=∫0∞1+1x2x−1x2+9=∫0∞dtt2+9=π6 where t=x−1x

If u=∫0∞dxx4+7x2+1 and v=∫0∞x2dxx4+7x2+1 then

u=∫0∞1x4+7x2+1dx Put x=1t,dx=−1t2=dtu=∫0∞11t4+7t2+1−dtt2=∫0∞t2dtt4+7t2+1=v⇒u=vSo, u+v=2u=∫0∞1+x2x4+7x2+1⇒2u=∫0∞1+1x2x−1x2+9=∫0∞dtt2+9=π6 where t=x−1x

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If $u = \int_{0}^{\infty} \frac{d x}{x^{4} + 7 x^{2} + 1}$ and $v = \int_{0}^{\infty} \frac{x^{2} d x}{x^{4} + 7 x^{2} + 1}$ then

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$u > v$

$u = v$

$u < v$

$u = \frac{π}{6}$

answer is C, D.

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Detailed Solution

$u = \int_{0}^{\infty} \frac{1}{x^{4} + 7 x^{2} + 1} d x$ Put $x = \frac{1}{t}, d x = \frac{- 1}{t^{2}} = d t$
$u = \int_{0}^{\infty} \frac{1}{\frac{1}{t^{4}} + \frac{7}{t^{2}} + 1} (\frac{- d t}{t^{2}}) = \int_{0}^{\infty} \frac{t^{2} d t}{t^{4} + 7 t^{2} + 1} = v$ $\Rightarrow u = v$

So, $u + v = 2 u = \int_{0}^{\infty} \frac{1 + x^{2}}{x^{4} + 7 x^{2} + 1}$ $\Rightarrow 2 u = \int_{0}^{\infty} \frac{(1 + \frac{1}{x^{2}})}{{(x - \frac{1}{x})}^{2} + 9} = \int_{0}^{\infty} \frac{d t}{t^{2} + 9} = \frac{π}{6}$
where $t = x - \frac{1}{x}$