Q41 / 75 · JEE Main 2025 (23 Jan Shift 1) · Mathematics
The value of
$\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$ is
- A$2$
- B$\log _e 2$
- Correct answerC$1$
- D$e^2$
Solution
$\begin{aligned}
& \text {Put } \ln x=t \Rightarrow \frac{1}{x} d x=d t \quad \begin{array}{|c|c|}
\hline x & t \\
\hline e^2 & 2 \\
\hline e^4 & 4 \\
\hline \end{array} \\
& I=\int_2^4 \frac{e^{\left(\mathrm{t}^2+1\right)^{-1}}}{e^{\left(\mathrm{t}^2+1\right)^{-1}}+e^{\left((6-t)^2+1\right)^{-1}}} d t \quad \ldots \text { (i) } \\
& I=\int_2^4 \frac{e^{\left((6-t)^2+1\right)^{-1}}}{e^{\left((6-t)^2+1\right)^{-1}}+e^{\left(t^2+1\right)^{-1}} d t} \ldots \text { (ii) } \\
& \left\{\text { Using } \int_a^0 f(x) d x=\int_a^0 f(a+b-x) d x\right\}
\end{aligned}$
Adding (i) and (ii) gives
$2 l=\int d t \Rightarrow l=1$
Hence, the correct answer is option: 'C'.