KCET · Maths · Definite Integration
\(\int_0^1 \log \left(\frac{1}{x}-1\right) d x\) is
- A \(1\)
- B \(0\)
- C \(\log _2 2\)
- D \(\log _e\left(\frac{1}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x \\ & I=\int_0^1 \log \left(\frac{1}{1+0-x}-1\right)=\int_0^1 \log \left(\frac{1}{1-x}-1\right) d x \\ & I=\int_0^1 \log \left(\frac{1-1+x}{1-x}\right) d x=\int_0^1 \log \left(\frac{x}{1-x}\right) d x \\ & 2 I=\int_0^1\left(\log \left(\frac{1-x}{x}\right)+\log \left(\frac{x}{1-x}\right)\right) d x \\ & 2 I=0 \quad \Rightarrow I=0 .\end{aligned}\)
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