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.

List $I$	List $II$
$P.$ The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$,satisfying $f(0)=0$ and $\int_0^1 f(x) dx=1$,is	$1.$ $8$
$Q.$ The number of points in the interval $(-\sqrt{13}, \sqrt{13})$ at which $f(x)=\sin(x^2)+\cos(x^2)$ attains its maximum value,is	$2.$ $2$
$R.$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx$ equals	$3.$ $4$
$S.$ $\frac{\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}{\int_0^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}$ equals	$4.$ $0$

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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