By substituting [tex]y=e^x+2[/tex], or [tex]x=\ln(y-2)[/tex] so that [tex]\mathrm dx=\frac{\mathrm dy}{y-2}[/tex], we have
[tex]\displaystyle\int\frac{\mathrm dx}{e^x+2}=\int\frac{\mathrm dy}{y(y-2)}[/tex]
Split the integrand into partial fractions:
[tex]\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)[/tex]
Then integrating gives
[tex]\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\frac12(\ln|y-2|-\ln|y|)+C[/tex]
Back-substitute to get
[tex]\displaystyle\int\frac{\mathrm dx}{e^x+2}=\frac12(\ln|e^x|-\ln|e^x+2|)+C=\frac{x-\ln(e^x+2)}2+C[/tex]
