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

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

Para a integral $$$\int{\ln{\left(x \right)}^{2} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

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

Então $$$\operatorname{du}=\left(\ln{\left(x \right)}^{2}\right)^{\prime }dx=\frac{2 \ln{\left(x \right)}}{x} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{1 d x}=x$$$ (os passos podem ser vistos »).

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=2$$$ e $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:

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

Para a integral $$$\int{\ln{\left(x \right)} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

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

Então $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{1 d x}=x$$$ (os passos podem ser vistos »).

Portanto,

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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