WBJEE · Maths · Definite Integration
\(\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=\)
- 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 \(1-\frac{3}{4} \log _{e} 3\)
Answer & Solution
Correct Answer
(B) \(1+\frac{3}{4} \log _{e} 3\)
Step-by-step Solution
Detailed explanation
\(\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=\int_{1}^{2} \frac{x-1}{-x+2-x+3} d x+\int_{2}^{3} \frac{x-1}{x-2-x+3} d x=\int_{1}^{2} \frac{x-1}{-2 x+5}+\int_{2}^{3} \frac{x-1}{1} d x=1+\frac{3}{4} \ln 3\)
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