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

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Answer 1 · 2014-09-20T16:47:01

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,

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

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