The value of int_{e^2}^{e^4} frac{1}{x} left( | vedclass

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

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(C) Let $I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$. Substitute $\ln x = t$,then $\frac{1}{x} dx = dt$. When $x = e^2$,$t = 2$. When $x = e^4$,$t = 4$. So,$I = \int_{2}^{4} \frac{e^{(t^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt$. Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,we get: $I = \int_{2}^{4} \frac{e^{((6-t)^2+1)^{-1}}}{e^{((6-t)^2+1)^{-1}} + e^{(t^2+1)^{-1}}} dt$. Adding the two expressions for $I$: $2I = \int_{2}^{4} \frac{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt = \int_{2}^{4} 1 dt = [t]_{2}^{4} = 4 - 2 = 2$. Therefore,$I = 1$.

The value of $\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$ is

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