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 $t=2x+1$
$=1/2[(2x+1)ln|2x+1|-(2x+1)+c]$

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