WBJEE · Maths · Definite Integration
If \(\int_{\log _{e} 2}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e} \frac{3}{2}\) then the value of \(x\) is
- A 1
- B \(\mathrm{e}^{2}\)
- C \(\log 4\)
- D \(\frac{1}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(C) \(\log 4\)
Step-by-step Solution
Detailed explanation
\(\int_{\log _{e}{ }^{2}}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e}^{\frac{3}{2}} \int_{\log _{e}{ }^{2}}^{x} \frac{e^{-x}}{1-e^{-x}} d x=\log _{e}^{\frac{3}{2}}\) \(e^{-x}=\frac{1}{4}\) \(x=\ln 4\)
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