Verified Solution • 2025–26 Session moderate

QJEEM19Madbf8: JEE Main 2019 Mathematics - Class 12 JEEM19Madbf8 Solution: The integral log e x dx is equal to :

The integral ∫1e{(xe)2x−(ex)x} \int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \, loge x dx is equal to :

Multiple Choice Options

  • A −12+1e−12e2- {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}
  • B 32−e−12e2{3 \over 2} - e - {1 \over {2{e^2}}}
  • C 12−e−1e2{1 \over 2} - e - {1 \over {{e^2}}}
  • D 32−1e−12x2{3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}

Step-by-Step Verified Solution

∫1e{(xe)2x−(ex)x} \int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,

Let   (xe)2x=t,(ex)x=v{\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v

=12∫(1e)21dt+∫e1dv= {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} }

=12(1−1e2)+(1−e)=32−12e2−e= {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e

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