How do I find the integral #intln(2x+1)dx# ?

By Substitution and Integration by Parts,

#int ln(2x+1)dx=1/2(2x+1)[ln(2x+1)-1]+C#

Let us look at some details.

#int ln(2x+1)dx#

by the substitution #t=2x+1#.
#Rightarrow {dt}/{dx}=2 Rightarrow {dx}/{dt}=1/2 Rightarrow dx={dt}/{2}#

#=1/2int ln t dt#

by Integration by Parts,
Let #u=ln t# and #dv=dt#
#Rightarrow du=dt/t# and #v=t#

#=1/2(tlnt-int dt)#

#=1/2(tlnt-t)+C#

by factoring out #t#,

#=1/2t(lnt-1)+C#

by putting #t=2x+1# back in,

#=1/2(2x+1)[ln(2x+1)-1]+C#