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

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

The input is rewritten: $$$\int{\ln{\left(x^{2} \right)}^{2} d x}=\int{4 \ln{\left(x \right)}^{2} d x}$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \ln{\left(x \right)}^{2}$$$:

$${\color{red}{\int{4 \ln{\left(x \right)}^{2} d x}}} = {\color{red}{\left(4 \int{\ln{\left(x \right)}^{2} d x}\right)}}$$

For the integral $$$\int{\ln{\left(x \right)}^{2} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(x \right)}^{2}$$$ and $$$\operatorname{dv}=dx$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x \right)}^{2}\right)^{\prime }dx=\frac{2 \ln{\left(x \right)}}{x} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen »).

Thus,

$$4 {\color{red}{\int{\ln{\left(x \right)}^{2} d x}}}=4 {\color{red}{\left(\ln{\left(x \right)}^{2} \cdot x-\int{x \cdot \frac{2 \ln{\left(x \right)}}{x} d x}\right)}}=4 {\color{red}{\left(x \ln{\left(x \right)}^{2} - \int{2 \ln{\left(x \right)} d x}\right)}}$$

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

$$4 x \ln{\left(x \right)}^{2} - 4 {\color{red}{\int{2 \ln{\left(x \right)} d x}}} = 4 x \ln{\left(x \right)}^{2} - 4 {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}$$

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

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

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

Thus,

$$4 x \ln{\left(x \right)}^{2} - 8 {\color{red}{\int{\ln{\left(x \right)} d x}}}=4 x \ln{\left(x \right)}^{2} - 8 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=4 x \ln{\left(x \right)}^{2} - 8 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$4 x \ln{\left(x \right)}^{2} - 8 x \ln{\left(x \right)} + 8 {\color{red}{\int{1 d x}}} = 4 x \ln{\left(x \right)}^{2} - 8 x \ln{\left(x \right)} + 8 {\color{red}{x}}$$

Therefore,

$$\int{4 \ln{\left(x \right)}^{2} d x} = 4 x \ln{\left(x \right)}^{2} - 8 x \ln{\left(x \right)} + 8 x$$

Simplify:

$$\int{4 \ln{\left(x \right)}^{2} d x} = 4 x \left(\ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} + 2\right)$$

Add the constant of integration:

$$\int{4 \ln{\left(x \right)}^{2} d x} = 4 x \left(\ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} + 2\right)+C$$

$$$\int \ln^{2}\left(x^{2}\right)\, dx = 4 x \left(\ln^{2}\left(x\right) - 2 \ln\left(x\right) + 2\right) + C$$$A