int_{1}^{3} frac{|x-1|}{|x-2|+|x-3|} d x= | vedclass

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(B) Let $I = \int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x$. Since $x \in [1, 3]$,$|x-1| = x-1$. For $x \in [1, 2]$,$|x-2| = 2-x$ and $|x-3| = 3-x$. For

Vedclass · Accepted Answer

$1+\frac{3}{4} \log _{e} 3$

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(B) Let $I = \int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x$. Since $x \in [1, 3]$,$|x-1| = x-1$. For $x \in [1, 2]$,$|x-2| = 2-x$ and $|x-3| = 3-x$. For $x \in [2, 3]$,$|x-2| = x-2$ and $|x-3| = 3-x$. Thus,$I = \int_{1}^{2} \frac{x-1}{(2-x)+(3-x)} d x + \int_{2}^{3} \frac{x-1}{(x-2)+(3-x)} d x$. $I = \int_{1}^{2} \frac{x-1}{5-2x} d x + \int_{2}^{3} (x-1) d x$. For the first integral,let $u = 5-2x$,then $du = -2 dx$,so $dx = -\frac{1}{2} du$. When $x=1, u=3$; when $x=2, u=1$. $\int_{1}^{2} \frac{x-1}{5-2x} d x = \int_{3}^{1} \frac{\frac{5-u}{2}-1}{u} (-\frac{1}{2}) du = \frac{1}{4} \int_{1}^{3} \frac{3-u}{u} du = \frac{1}{4} [3 \ln |u| - u]_{1}^{3} = \frac{1}{4} (3 \ln 3 - 2) = \frac{3}{4} \ln 3 - \frac{1}{2}$. For the second integral,$\int_{2}^{3} (x-1) d x = [\frac{x^2}{2} - x]_{2}^{3} = (\frac{9}{2} - 3) - (2 - 2) = \frac{3}{2}$. Therefore,$I = (\frac{3}{4} \ln 3 - \frac{1}{2}) + \frac{3}{2} = 1 + \frac{3}{4} \ln 3$.

$\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=$

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