Let #e^x=u#, #=>#, #du=e^xdx#
Therefore, the integral is
#I=int(dx)/(e^x(3e^x+2))=int(du)/(e^(2x)(3e^x+2))#
#=int(du)/(u^2(3u+2))#
Perform the decomposition into partial fractions
#1/(u^2(3u+2))=A/u^2+B/u+C/(3u+2)#
#=(A(3u+2)+Bu(3u+2)+Cu^2)/(u^2(3u+2))#
The denominator is the same, compare the numerators
#1=A(3u+2)+Bu(3u+2)+Cu^2#
Let #u=0#, #=>#, #1=2A#, #=>#, #A=1/2#
Coefficients of #u#
#0=3A+2B#, #=>#, #B=-3/2A=-3/4#
Coefficients of #u^2#
#0=3B+C#, #=>#, #C=-3B=9/4#
Therefore,
#1/(u^2(3u+2))=(1/2)/u^2+(-3/4)/u+(9/4)/(3u+2)#
So,
#int(du)/(u^2(3u+2))=1/2int(du)/u^2-3/4int(du)/u+9/4int(du)/(3u+2)#
#=-1/(2u)-3/4lnu+3/4ln(3u+2)#
#=-1/(2e^x)-3/4ln(e^x)+3/4ln(3e^x+2)+C#
#=-1/(2e^x)-3/4x+3/4ln(3e^x+2)+C#