Q85 / 155 · WBJEE 2021 · Mathematics

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

  • A
    $1+\frac{4}{3} \log _{e} 3$
  • Correct answer
    B
    $1+\frac{3}{4} \log _{e} 3$
  • C
    $1-\frac{4}{3} \log _{e} 3$
  • D
    $1-\frac{3}{4} \log _{e} 3$

Solution

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

Hence, the correct answer is option: 'B'.

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