1. Write the integral clearly

We need to evaluate

$$I=\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left((\ln x)^2+1\right)^{-1}}}{e^{\left((\ln x)^2+1\right)^{-1}}+e^{\left((6-\ln x)^2+1\right)^{-1}}}\right)dx.$$

2. Substitute $t=\ln x$

Since

$$t=\ln x \quad\Rightarrow\quad dt=\frac{dx}{x},$$

and the limits change as:

$$x=e^2 \Rightarrow t=2, \qquad x=e^4 \Rightarrow t=4.$$

So the integral becomes

$$I=\int_2^4 \frac{e^{\frac{1}{t^2+1}}}{e^{\frac{1}{t^2+1}}+e^{\frac{1}{(6-t)^2+1}}}\,dt.$$

3. Observe the symmetry

Define

$$f(t)=\frac{e^{\frac{1}{t^2+1}}}{e^{\frac{1}{t^2+1}}+e^{\frac{1}{(6-t)^2+1}}}.$$

Now compute $f(6-t)$:

$$f(6-t)=\frac{e^{\frac{1}{(6-t)^2+1}}}{e^{\frac{1}{(6-t)^2+1}}+e^{\frac{1}{t^2+1}}}.$$

Thus,

$$f(t)+f(6-t)=1.$$

Also, the interval $[2,4]$ is symmetric about $t=3$, and under the transformation $u=6-t$, the interval $[2,4]$ maps onto itself.

Therefore,

$$\int_2^4 f(t)\,dt = \int_2^4 f(6-t)\,dt.$$

Adding these,

$$2I=\int_2^4 \big(f(t)+f(6-t)\big)\,dt=\int_2^4 1\,dt=4-2=2.$$

Hence,

$$I=1.$$

4. Check options

• A: $1$ ✅
• B: $\ln 2$ ❌
• C: $e^2$ ❌
• D: $2$ ❌

So the correct option is A.