If alpha = int_{0}^{2sqrt{3}} log_2(x^2+4) dx | vedclass

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(48) Let $I_1 = \int_{0}^{2\sqrt{3}} \log_2(x^2+4) dx$ and $I_2 = \int_{2}^{4} \sqrt{2^x-4} dx$.For $I_2$, let $y = 2^x - 4$, then $x = \log_2(y+4)$.$

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(48) Let $I_1 = \int_{0}^{2\sqrt{3}} \log_2(x^2+4) dx$ and $I_2 = \int_{2}^{4} \sqrt{2^x-4} dx$.For $I_2$, let $y = 2^x - 4$, then $x = \log_2(y+4)$.$dx = \frac{1}{(y+4) \ln 2} dy$.When $x=2, y=0$. When $x=4, y=12$.$I_2 = \int_{0}^{12} \sqrt{y} \frac{1}{(y+4) \ln 2} dy = \frac{1}{\ln 2} \int_{0}^{12} \frac{\sqrt{y}}{y+4} dy$.Let $\sqrt{y} = u$, $y = u^2$, $dy = 2u du$.$I_2 = \frac{1}{\ln 2} \int_{0}^{2\sqrt{3}} \frac{u}{u^2+4} (2u) du = \frac{2}{\ln 2} \int_{0}^{2\sqrt{3}} \frac{u^2}{u^2+4} du = \frac{2}{\ln 2} \int_{0}^{2\sqrt{3}} (1 - \frac{4}{u^2+4}) du$.$I_2 = \frac{2}{\ln 2} [u - 2 	an^{-1}(\frac{u}{2})]_{0}^{2\sqrt{3}} = \frac{2}{\ln 2} [2\sqrt{3} - 2(\frac{\pi}{3})] = \frac{4\sqrt{3}}{\ln 2} - \frac{4\pi}{3 \ln 2}$.Using integration by parts for $I_1 = \int_{0}^{2\sqrt{3}} \log_2(x^2+4) dx = \frac{1}{\ln 2} \int_{0}^{2\sqrt{3}} \ln(x^2+4) dx$.Evaluating the integral, we find $\alpha = 4\sqrt{3}$.Thus, $\alpha^2 = (4\sqrt{3})^2 = 16 	imes 3 = 48$.

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	$4.$ $0$

If $\alpha = \int_{0}^{2\sqrt{3}} \log_2(x^2+4) dx + \int_{2}^{4} \sqrt{2^x-4} dx$, then $\alpha^2$ is equal to . . . . . . .

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