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

Sep 20, 2014

By Substitution and Integration by Parts,

$\int \ln \left(2 x + 1\right) \mathrm{dx} = \frac{1}{2} \left(2 x + 1\right) \left[\ln \left(2 x + 1\right) - 1\right] + C$

Let us look at some details.

$\int \ln \left(2 x + 1\right) \mathrm{dx}$

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

$= \frac{1}{2} \int \ln t \mathrm{dt}$

by Integration by Parts,
Let $u = \ln t$ and $\mathrm{dv} = \mathrm{dt}$
$R i g h t a r r o w \mathrm{du} = \frac{\mathrm{dt}}{t}$ and $v = t$

$= \frac{1}{2} \left(t \ln t - \int \mathrm{dt}\right)$

$= \frac{1}{2} \left(t \ln t - t\right) + C$

by factoring out $t$,

$= \frac{1}{2} t \left(\ln t - 1\right) + C$

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

$= \frac{1}{2} \left(2 x + 1\right) \left[\ln \left(2 x + 1\right) - 1\right] + C$