"I am Back" Says Integration Part 10

$\begin{aligned} I & = \int_1^e \left[\left(\frac xe\right)^x + \left(\frac ex\right)^x \right] \ln x \ dx \\ & = \int_1^e \left[e^{\color{#3D99F6}x(\ln x - 1)} + e^{-\color{#3D99F6}x(\ln x - 1)} \right] \ln x \ dx & \small \color{#3D99F6} \text{Let }u = x(\ln x -1), \ du = \ln x \ dx \\ & = \int_{\color{#3D99F6}-1} ^{\color{#3D99F6}0} \left[e^{\color{#3D99F6}u} + e^{-\color{#3D99F6}u} \right] \ d\color{#3D99F6}u \\ & = e^u - e^{-u} \bigg|_{-1}^0 \\ & = 1-1-e^{-1} + e^1 \\ & = e-\frac 1e \end{aligned}$

$\implies a + b = 1-1 = \boxed{0}$