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$\int_{\log 2}^{k} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6} then k =$

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Correct option is C

∫log⁡2k exdxex⋅ex−1 put ex−1=t∫1e−1 2tdtt2+1t ex−1=t2exdx=2tdt2∫1eε−1 dt1+t2=2tan−1⁡t1et−1 LLt=2−1t=1∣ULt=ek−1⇒2tan−1⁡ek−1−π4=π62tan−1⁡ek−1=π6+π2tan−1⁡ek−1=π3ek−1=3ek=4k=log⁡4

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∫log⁡2k dxex−1=π6 then k=

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$\int_{\log 2}^{k} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6} then k =$