Question

$$
\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=
$$
(A) \(1+\frac{4}{3} \log _{e} 3\)
(B) \(1+\frac{3}{4} \log _{e} 3\)
(C) \(1-\frac{4}{3} \log _{e} 3\)
(D) \(\quad 1-\frac{3}{4} \log _{e} 3\)

Answer 1

Ans: (B)
Hint \(: \int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=\int_{1}^{2} \frac{x-1}{-x+2-x+3} d x+\int_{2}^{3} \frac{x-1}{x-2-x+3} d x=\int_{1}^{2} \frac{x-1}{-2 x+5}+\int_{2}^{3} \frac{x-1}{1} d x=1+\frac{3}{4} \ln 3\)