int_3^8 frac{2 - 3x}{xsqrt{1 + x}} , dx is eq | vedclass

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(A) Let $I = \int_3^8 \frac{2 - 3x}{x\sqrt{1 + x}} \, dx$.Substitute $1 + x = t^2$,which implies $x = t^2 - 1$ and $dx = 2t \, dt$.When $x = 3$,$t = \

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$2\log \left( \frac{3}{2e^3} \right)$

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(A) Let $I = \int_3^8 \frac{2 - 3x}{x\sqrt{1 + x}} \, dx$.Substitute $1 + x = t^2$,which implies $x = t^2 - 1$ and $dx = 2t \, dt$.When $x = 3$,$t = \sqrt{1 + 3} = 2$. When $x = 8$,$t = \sqrt{1 + 8} = 3$.Substituting these into the integral:$I = \int_2^3 \frac{2 - 3(t^2 - 1)}{(t^2 - 1)t} \cdot 2t \, dt = 2 \int_2^3 \frac{5 - 3t^2}{t^2 - 1} \, dt$.$I = 2 \int_2^3 \left( \frac{2}{t^2 - 1} - 3 \right) \, dt$.Using the formula $\int \frac{1}{t^2 - a^2} \, dt = \frac{1}{2a} \log \left| \frac{t - a}{t + a} \right|$,we get:$I = 2 \left[ \frac{2}{2(1)} \log \left| \frac{t - 1}{t + 1} \right| - 3t \right]_2^3$.$I = 2 \left[ \log \left| \frac{t - 1}{t + 1} \right| - 3t \right]_2^3$.Evaluating at the limits:$I = 2 \left[ (\log(2/4) - 9) - (\log(1/3) - 6) \right] = 2 \left[ \log(1/2) - \log(1/3) - 3 \right]$.$I = 2 \left[ \log(3/2) - 3 \right] = 2 \log \left( \frac{3}{2e^3} \right)$.

$\int_3^8 \frac{2 - 3x}{x\sqrt{1 + x}} \, dx$ is equal to

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