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

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

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

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

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

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

So,

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

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

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

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

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

Then $$$\operatorname{dg}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (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 $$$u=- x$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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