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Question Number 17475 by Arnab Maiti last updated on 06/Jul/17
Prove that ∫_e ^e^2  ((lnx)/((1+lnx)^2 ))dx=(e/6)(2e−3)
$$\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{e}} ^{\mathrm{e}^{\mathrm{2}} } \frac{\mathrm{ln}{x}}{\left(\mathrm{1}+\mathrm{ln}{x}\right)^{\mathrm{2}} }\mathrm{d}{x}=\frac{\mathrm{e}}{\mathrm{6}}\left(\mathrm{2e}−\mathrm{3}\right) \\ $$
Answered by ajfour last updated on 06/Jul/17
lnx=t   ⇒   (dx/x)=dt  or  dx=e^t dt  t=2 when x=e^2 ,  t=1  for  x=e  I=∫_e ^(  e^2 ) ((ln x)/((1+ln x)^2 ))dx=∫_1 ^(  2) ((te^t )/((1+t)^2 ))dt     =∫_1 ^(  2) e^t [(1/(1+t))−(1/((1+t)^2 ))]dt     =((e^t /(1+t)))∣_1 ^2  =(e^2 /3)−(e^2 /2) =((e(2e−3))/6) .
$$\mathrm{lnx}=\mathrm{t}\:\:\:\Rightarrow\:\:\:\frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\:\:\mathrm{or}\:\:\mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt} \\ $$$$\mathrm{t}=\mathrm{2}\:\mathrm{when}\:\mathrm{x}=\mathrm{e}^{\mathrm{2}} ,\:\:\mathrm{t}=\mathrm{1}\:\:\mathrm{for}\:\:\mathrm{x}=\mathrm{e} \\ $$$$\mathrm{I}=\int_{\mathrm{e}} ^{\:\:\mathrm{e}^{\mathrm{2}} } \frac{\mathrm{ln}\:\mathrm{x}}{\left(\mathrm{1}+\mathrm{ln}\:\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}=\int_{\mathrm{1}} ^{\:\:\mathrm{2}} \frac{\mathrm{te}^{\mathrm{t}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:\:=\int_{\mathrm{1}} ^{\:\:\mathrm{2}} \mathrm{e}^{\mathrm{t}} \left[\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}}−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\right]\mathrm{dt} \\ $$$$\:\:\:=\left(\frac{\mathrm{e}^{\mathrm{t}} }{\mathrm{1}+\mathrm{t}}\right)\mid_{\mathrm{1}} ^{\mathrm{2}} \:=\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{2}}\:=\frac{\boldsymbol{\mathrm{e}}\left(\mathrm{2}\boldsymbol{\mathrm{e}}−\mathrm{3}\right)}{\mathrm{6}}\:. \\ $$

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