Solve [(alpha)^2]u''+a*e^u=0 by multiplying it by du/dx and integrating it in x
I just keep going in circles and ending up with the original problem.
$\displaystyle \displaystyle \begin{align*} \alpha ^2 \, \frac{d^2u}{dx^2} + a\, e^u &= 0 \\ \alpha ^2 \, \frac{d^2u}{dx^2} &= -a\, e^u \\ \alpha ^2 \, \frac{du}{dx} \, \frac{d^2u}{dx^2} &= -a\, e^u \, \frac{du}{dx} \end{align*}$
Now, if we let $\displaystyle \displaystyle \begin{align*} w = \frac{du}{dx} \end{align*}$, then $\displaystyle \displaystyle \begin{align*} \frac{dw}{dx} = \frac{d^2u}{dx^2} \end{align*}$ and the DE becomes
$\displaystyle \displaystyle \begin{align*} \alpha ^2 \, w \, \frac{dw}{dx} &= -a \, e^u \, \frac{du}{dx} \\ \int{ \alpha^2 \, w \, \frac{dw}{dx} \, dx} &= \int{ -a\, e^u \, \frac{du}{dx} \, dx} \\ \int{\alpha^2 \, w \, dw} &= \int{ -a \, e^u \, du} \end{align*}$
Finish it.