Integral of $$$\ln\left(z^{2}\right)$$$

Find $$$\int \ln\left(z^{2}\right)\, dz$$$.

The input is rewritten: $$$\int{\ln{\left(z^{2} \right)} d z}=\int{2 \ln{\left(z \right)} d z}$$$.

Apply the constant multiple rule $$$\int c f{\left(z \right)}\, dz = c \int f{\left(z \right)}\, dz$$$ with $$$c=2$$$ and $$$f{\left(z \right)} = \ln{\left(z \right)}$$$:

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

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

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

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

Thus,

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

Apply the constant rule $$$\int c\, dz = c z$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \ln\left(z^{2}\right)\, dz = 2 z \left(\ln\left(z\right) - 1\right) + C$$$A