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

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

Let $$$u=\frac{1}{x}$$$.

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

The integral can be rewritten as

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

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

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

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

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

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

The integral can be rewritten as

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

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

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

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

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

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

The integral becomes

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Recall that $$$u=\frac{1}{x}$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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