intlimits_0^3 left( frac{1}{sqrt{x^2 + 4x + 4 | vedclass

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Question

(C) The integral is $I = \int_0^3 \left( \frac{1}{\sqrt{(x+2)^2}} + \sqrt{(x-2)^2} \right) dx = \int_0^3 \left( \frac{1}{|x+2|} + |x-2| \right) dx$. S

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$\ln \frac{5}{2} + \frac{5}{2}$

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(C) The integral is $I = \int_0^3 \left( \frac{1}{\sqrt{(x+2)^2}} + \sqrt{(x-2)^2} \right) dx = \int_0^3 \left( \frac{1}{|x+2|} + |x-2| \right) dx$. Since $x \in [0, 3]$,$|x+2| = x+2$. Thus,$I = \int_0^3 \frac{1}{x+2} dx + \int_0^3 |x-2| dx$. Evaluating the first part: $\int_0^3 \frac{1}{x+2} dx = [\ln|x+2|]_0^3 = \ln 5 - \ln 2 = \ln \frac{5}{2}$. Evaluating the second part: $\int_0^3 |x-2| dx = \int_0^2 -(x-2) dx + \int_2^3 (x-2) dx$. $= [-\frac{x^2}{2} + 2x]_0^2 + [\frac{x^2}{2} - 2x]_2^3$. $= (-2 + 4) + ((\frac{9}{2} - 6) - (2 - 4)) = 2 + (-\frac{3}{2} + 2) = 2 + \frac{1}{2} = \frac{5}{2}$. Therefore,$I = \ln \frac{5}{2} + \frac{5}{2}$.

$\int\limits_0^3 \left( \frac{1}{\sqrt{x^2 + 4x + 4}} + \sqrt{x^2 - 4x + 4} \right) dx =$

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