Let u = int_{0}^{infty} frac{dx}{x^4 + 7x^2 + | vedclass

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(B) For $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$,substitute $x = 1/t$,so $dx = -1/t^2 dt$. As $x 	o 0, t 	o \infty$ and as $x 	o \infty, t

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$6v = \pi$

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(B) For $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$,substitute $x = 1/t$,so $dx = -1/t^2 dt$. As $x 	o 0, t 	o \infty$ and as $x 	o \infty, t 	o 0$. $u = \int_{\infty}^{0} \frac{-1/t^2 dt}{1/t^4 + 7/t^2 + 1} = \int_{0}^{\infty} \frac{t^2 dt}{1 + 7t^2 + t^4} = v$. Thus,$u = v$. Now,consider $u + v = \int_{0}^{\infty} \frac{1 + x^2}{x^4 + 7x^2 + 1} dx = \int_{0}^{\infty} \frac{1 + 1/x^2}{x^2 + 7 + 1/x^2} dx$. Let $x - 1/x = z$,then $(1 + 1/x^2) dx = dz$. When $x 	o 0, z 	o -\infty$ and when $x 	o \infty, z 	o \infty$. $u + v = \int_{-\infty}^{\infty} \frac{dz}{z^2 + 2 + 7} = \int_{-\infty}^{\infty} \frac{dz}{z^2 + 9} = \left[ \frac{1}{3} 	an^{-1} \left( \frac{z}{3} ight) ight]_{-\infty}^{\infty} = \frac{1}{3} (\pi/2 - (-\pi/2)) = \pi/3$. Since $u = v$,$2v = \pi/3 \Rightarrow v = \pi/6$. Checking options: $(A)$ $v > u$ is false since $u = v$. $(B)$ $6v = 6(\pi/6) = \pi$,which is true. $(C)$ $3u + 2v = 3(\pi/6) + 2(\pi/6) = 5\pi/6$,which is true. Since $(B)$ and $(C)$ are true,the question structure implies a potential error in the options provided,but based on the calculation,$(B)$ and $(C)$ are correct.

Column $I$	Column $II$
$(A) \int_{-1}^1 \frac{dx}{1+x^2}$	$(p) \frac{1}{2} \log \left(\frac{2}{3}\right)$
$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}}$	$(q) 2 \log \left(\frac{2}{3}\right)$
$(C) \int_2^3 \frac{dx}{1-x^2}$	$(r) \frac{\pi}{3}$
$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}}$	$(s) \frac{\pi}{2}$

Let $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$ and $v = \int_{0}^{\infty} \frac{x^2 dx}{x^4 + 7x^2 + 1}$. Then:

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