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$\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=$
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$1+\frac{3}{4} \log _{e} 3$
$\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=\int_{1}^{2} \frac{x-1}{-x+2-x+3} d x+\int_{2}^{3} \frac{x-1}{x-2-x+3} d x=\int_{1}^{2} \frac{x-1}{-2 x+5}+\int_{2}^{3} \frac{x-1}{1} d x=1+\frac{3}{4} \ln 3$
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