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

Let $u=2 x + 1$.

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

Thus,

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

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 $\int \operatorname{q} \operatorname{dv} = \operatorname{q}\operatorname{v} - \int \operatorname{v} \operatorname{dq}$.

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

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

The integral can be rewritten as

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

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