The value of $\displaystyle\int_{e^2}^{e^4}\frac{1}{x}\left(\frac{e^{(\log_e x)^2+1}-1}{e^{(\log_e x)^2+1}-1+e^{(6-\log_e x)^2+1}-1}\right)dx$ is
Step-by-Step Solution
Key Concept: Substitute $t = \ln x$ so the limits become $2$ to $4$ and $dx/x = dt$; then apply the symmetry property $\int_a^b f(t)dt = \int_a^b f(a+b-t)dt$ with $a+b=6$ to get $2I = \int_2^4 dt = 2$.
Let $t = \ln x$, $dt = dx/x$. Limits: $x=e^2\to t=2$, $x=e^4\to t=4$.
$$I = \int_2^4\frac{e^{(t^2+1)^{-1}}-1}{e^{(t^2+1)^{-1}}-1+e^{((6-t)^2+1)^{-1}}-1}\,dt \quad\cdots(1)$$
Replace $t\to 6-t$ (since $a+b=2+4=6$):
$$I = \int_2^4\frac{e^{((6-t)^2+1)^{-1}}-1}{e^{((6-t)^2+1)^{-1}}-1+e^{(t^2+1)^{-1}}-1}\,dt \quad\cdots(2)$$
Adding $(1)+(2)$: $2I = \int_2^4 1\,dt = 2 \Rightarrow I = 1$.
Correct Answer: 3