Integral of $$$\ln\left(2 x + 1\right)$$$

Find $$$\int \ln\left(2 x + 1\right)\, dx$$$.

Let $$$u=2 x + 1$$$.

Then $$$du=\left(2 x + 1\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

So,

$${\color{red}{\int{\ln{\left(2 x + 1 \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{2}\right)}}$$

For the integral $$$\int{\ln{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{o} \operatorname{dv} = \operatorname{o}\operatorname{v} - \int \operatorname{v} \operatorname{do}$$$.

Let $$$\operatorname{o}=\ln{\left(u \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{do}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

So,

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{2}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{2}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{2}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$\frac{u \ln{\left(u \right)}}{2} - \frac{{\color{red}{\int{1 d u}}}}{2} = \frac{u \ln{\left(u \right)}}{2} - \frac{{\color{red}{u}}}{2}$$

Recall that $$$u=2 x + 1$$$:

$$- \frac{{\color{red}{u}}}{2} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{2} = - \frac{{\color{red}{\left(2 x + 1\right)}}}{2} + \frac{{\color{red}{\left(2 x + 1\right)}} \ln{\left({\color{red}{\left(2 x + 1\right)}} \right)}}{2}$$

Therefore,

$$\int{\ln{\left(2 x + 1 \right)} d x} = - x + \frac{\left(2 x + 1\right) \ln{\left(2 x + 1 \right)}}{2} - \frac{1}{2}$$

Simplify:

$$\int{\ln{\left(2 x + 1 \right)} d x} = \frac{\left(2 x + 1\right) \left(\ln{\left(2 x + 1 \right)} - 1\right)}{2}$$

Add the constant of integration:

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

$$$\int \ln\left(2 x + 1\right)\, dx = \frac{\left(2 x + 1\right) \left(\ln\left(2 x + 1\right) - 1\right)}{2} + C$$$A