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$\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x$ is equal to
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Verified Answer
The correct answer is:
$\log \frac{32}{27}$
Let $I=\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x$
$\quad=\int_{5}^{10}\left[\frac{-1}{x-1}+\frac{1}{x-2}\right] d x$
$=[-\log (x-1)+\log (x-2)]_{5}^{10}$
$=-\log 9+\log 8+\log 4-\log 3$
$=-2 \log 3+3 \log 2+2 \log 2-\log 3$
$=-3 \log 3+5 \log 2$
$=-\log 27+\log 32$
$=\log \frac{32}{27}$
$\quad=\int_{5}^{10}\left[\frac{-1}{x-1}+\frac{1}{x-2}\right] d x$
$=[-\log (x-1)+\log (x-2)]_{5}^{10}$
$=-\log 9+\log 8+\log 4-\log 3$
$=-2 \log 3+3 \log 2+2 \log 2-\log 3$
$=-3 \log 3+5 \log 2$
$=-\log 27+\log 32$
$=\log \frac{32}{27}$
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