Question

Let $u=\stackrel{\infty }{\underset{0}{\int }}\frac{dx}{x^4+7x^2+1}\&v=\stackrel{\infty }{\underset{0}{\int }}\frac{x^{2}dx}{x^4+7x^2+1}$ then:

• A. v > u
• B. 6v = $\pi$
• C. $3u+2v=5\pi /6$
• D. $u+v=\pi /3$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $3u+2v=5\pi /6$
C 6v = $\pi$
D $u+v=\pi /3$

$v={\int }_0^{\infty }\frac{x^{2}dx}{x^4+7x^2+1}$ put $x=\frac{1}{t}\Rightarrow dx=\frac{-1}{t^2}dt$
$v=-{\int }_{\infty }^0\frac{\frac{1}{t^2}\cdot \frac{1}{t^2}dt}{\frac{1}{t^4}+\frac{1}{t^2}+1}={\int }_0^{\infty }\frac{dx}{x^4+7x^2+1}$
v=u hence $2u={\int }_0^{\infty }\left(\frac{x^2+1}{x^4+7x^2+1}\right)dx$
$={\int }_0^{\infty }\left(\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+7}\right)dx={\int }_0^{\infty }\frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+3^2}={\int }_0^{\infty }\frac{dt}{t^2+9}$
$\frac{2}{3}{\left[{\tan }^{-1}\frac{t}{3}\right]}_0^{\infty }$
$2u=\pi /3$