After setting [tex]u=\sqrt{1+e^{2x}}[/tex], partially solving for x in terms of u gives
[tex]u = sqrt{1+e^{2x}} \implies u^2 = 1 + e^{2x} \implies e^{2x} = u^2 - 1[/tex]
Then taking differentials, you get
[tex]2 e^{2x} \,\mathrm dx = 2u \, \mathrm du \implies \mathrm dx = \dfrac{u}{u^2-1}\,\mathrm du[/tex]
Replacing everything in the original integral then gives
[tex]\displaystyle \int \frac9{\sqrt{1+e^{2x}}}\,\mathrm dx = \int \frac9u \times \frac u{u^2-1}\,\mathrm du = 9 \int \frac{\mathrm du}{u^2-1}[/tex]
Split up the integrand into partial fractions:
[tex]\dfrac1{u^2-1} = \dfrac a{u-1} + \dfrac b{u+1} \\\\ 1 = a(u+1) + b(u-1) = (a+b)u + a-b \\\\ \implies \begin{cases}a+b=0\\a-b=1\end{cases} \implies a=\dfrac12,b=-\dfrac12[/tex]
so that
[tex]\displaystyle 9 \int \frac{\mathrm du}{u^2-1} = \frac92 \int \left(\frac1{u-1} - \frac1{u+1}\right) \,\mathrm du \\\\ = \frac92 \left(\ln|u-1| - \ln|u+1|\right) + C \\\\ = \frac92 \ln\left|\frac{u-1}{u+1}\right| + C \\\\ = \frac92 \ln\left(\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\right) + C[/tex]
