NIMCET 2019 Question Paper with solutions

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Question : 28 of 120

Marks: +1, -0

If

∫↖{x}↙{\log 2} 1/ {√{e^x -1}

dx =

π/6

then x =

log 2
2 log 2
3 log 2
4 log 2

Solution:

Let

I=∫↖{x}↙{\log\;2}1/√{e^x-1}dx

Put

e^x-1=t^2⇒e^x\;dx=2t\;dt

⇒dx={2t}/{t^2+1}dt

∴I=∫↙{√{e^x-1}}↖{√{e^x-1}}{2t}/{(t^2+1)t}dt

=2∫↖{√{e^x-1}}↙{1}{dt}/{t^2+1}=2[\tan^{-1}\;t]^{√{e^x-1}}_1

∴\;\;2[\tan^{-1}√{e^{x}-1}-\tan^{-1}1]=π/6

⇒\tan^{-1}√{e^{x}-1}=\;\;{π}/{12}+\;\;{π}/{4}=\;\;{π}/{3}

⇒√{e^{x}-1}=\tan\;\;{π}/{3}=√{3}

⇒e^{x}-1=3⇒e^{x}=4

x=\log4=2\log2

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