$\begin{aligned}
& \text {Put } \ln x=t \Rightarrow \frac{1}{x} d x=d t \quad \begin{array}{|c|c|}
\hline x & t \
\hline e^2 & 2 \
\hline e^4 & 4 \
\hline \end{array} \
& I=\int_2^4 \frac{e^{\left(\mathrm{t}^2+1\right)^{-1}}}{e^{\left(\mathrm{t}^2+1\right)^{-1}}+e^{\left((6-t)^2+1\right)^{-1}}} d t \quad \ldots \text { (i) } \
& I=\int_2^4 \frac{e^{\left((6-t)^2+1\right)^{-1}}}{e^{\left((6-t)^2+1\right)^{-1}}+e^{\left(t^2+1\right)^{-1}} d t} \ldots \text { (ii) } \
& \left{\text { Using } \int_a^0 f(x) d x=\int_a^0 f(a+b-x) d x\right}
\end{aligned}Adding(i)and(ii)gives2 l=\int d t \Rightarrow l=1$