Q115 / 180 · KCET 2025 · Mathematics

$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ is

  • A
    $1$
  • Correct answer
    B
    $0$
  • C
    $\log _2 2$
  • D
    $\log _e\left(\frac{1}{2}\right)$

Solution

$\begin{aligned} & \int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x \\ & I=\int_0^1 \log \left(\frac{1}{1+0-x}-1\right)=\int_0^1 \log \left(\frac{1}{1-x}-1\right) d x \\ & I=\int_0^1 \log \left(\frac{1-1+x}{1-x}\right) d x=\int_0^1 \log \left(\frac{x}{1-x}\right) d x \\ & 2 I=\int_0^1\left(\log \left(\frac{1-x}{x}\right)+\log \left(\frac{x}{1-x}\right)\right) d x \\ & 2 I=0 \quad \Rightarrow I=0 .\end{aligned}$

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

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