Let I(x) = int sqrt{frac{x+7}{x}} , dx and I( | vedclass

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(D) Let $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$. Substitute $x = t^2$,then $dx = 2t \, dt$. The integral becomes $\int \sqrt{\frac{t^2+7}{t^2}} \cdot

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$64$

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(D) Let $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$. Substitute $x = t^2$,then $dx = 2t \, dt$. The integral becomes $\int \sqrt{\frac{t^2+7}{t^2}} \cdot 2t \, dt = 2 \int \sqrt{t^2+7} \, dt$. Using the formula $\int \sqrt{t^2+a^2} \, dt = \frac{t}{2} \sqrt{t^2+a^2} + \frac{a^2}{2} \ln|t + \sqrt{t^2+a^2}| + C$,we get: $I(t) = 2 \left[ \frac{t}{2} \sqrt{t^2+7} + \frac{7}{2} \ln|t + \sqrt{t^2+7}| \right] + C = t \sqrt{t^2+7} + 7 \ln|t + \sqrt{t^2+7}| + C$. Substituting $t = \sqrt{x}$,we get $I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln|\sqrt{x} + \sqrt{x+7}| + C$. Given $I(9) = 12 + 7 \ln 7$. $I(9) = \sqrt{9} \sqrt{9+7} + 7 \ln|\sqrt{9} + \sqrt{9+7}| + C = 3 \cdot 4 + 7 \ln(3+4) + C = 12 + 7 \ln 7 + C$. Comparing,we get $C = 0$. So,$I(x) = \sqrt{x(x+7)} + 7 \ln(\sqrt{x} + \sqrt{x+7})$. Now,$I(1) = \sqrt{1(1+7)} + 7 \ln(\sqrt{1} + \sqrt{1+7}) = \sqrt{8} + 7 \ln(1 + \sqrt{8}) = \sqrt{8} + 7 \ln(1 + 2\sqrt{2})$. Given $I(1) = \alpha + 7 \ln(1 + 2\sqrt{2})$,we find $\alpha = \sqrt{8}$. Therefore,$\alpha^4 = (\sqrt{8})^4 = 8^2 = 64$.

Let $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$ and $I(9) = 12 + 7 \log_e 7$. If $I(1) = \alpha + 7 \log_e(1 + 2\sqrt{2})$,then $\alpha^4$ is equal to $..........$.

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