Integral of $$$\ln\left(- 2 x\right)$$$

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

Let $$$u=- 2 x$$$.

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

Therefore,

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

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 $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

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

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

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

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

The integral becomes

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

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 $$$u=- 2 x$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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