How do you integrate #intln(2x+1)dx#?

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Answer 1 · 2015-05-07T00:41:05

I would set #2x+1=t#
#2dx=dt#
#intln(t)dt/2=# by parts:
#=1/2[tln(t)-intt*1/tdt]=#
#=1/2[tln(t)-t+c]#
but #t=2x+1#
#=1/2[(2x+1)ln|2x+1|-(2x+1)+c]#