Question

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) \(e^{2}\)
(C) \(\log 4\)
(D) \(\frac{1}{\mathrm{e}}\)

Answer 1

Ans : (C)
Hint \(: \int_{\log _{e}{ }^{2}}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e}{ }^{\frac{3}{2}} \int_{\log _{e}{ }^{2}}^{x} \frac{e^{-x}}{1-e^{-x}} d x=\log _{e} \frac{\frac{3}{2}}{2}\)
$$
\mathrm{e}^{-\mathrm{x}}=\frac{1}{4} \mathrm{x}=\ln 4
$$

Answer 2

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Answer 3

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