Verified Solution • 2025–26 Session moderate

QJEEM25Ma7fc5: JEE Main 2025 Mathematics - Class 12 JEEM25Ma7fc5 Solution

The value of ∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dx\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x is

Multiple Choice Options

  • A 1
  • B log⁡e2\log_e2
  • C e2e^2
  • D 2

Step-by-Step Verified Solution

amp; Let ln⁡x=t⇒dxx=dtamp;amp;I=∫24e11+t2e11+t2+e11+(6−t)2dtamp;I=∫241e1+(6−t)21e1+(6−t)2+e11+t2dtamp;2I=∫24dt=(t)24=4−2=2amp;I=1\begin{aligned}&\text { Let } \ln \mathrm{x}=\mathrm{t} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\\&\begin{aligned}& I=\int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}}+e^{\frac{1}{1+(6-t)^2}}} d t \\& I=\int_2^4 \frac{\frac{1}{e^{1+(6-t)^2}}}{\frac{1}{e^{1+(6-t)^2}}+e^{\frac{1}{1+t^2}}} d t \\& 2 I=\int_2^4 d t=(t)_2^4=4-2=2 \\& I=1\end{aligned}\end{aligned}

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