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

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

A entrada é reescrita como: $$$\int{\frac{\ln{\left(x^{2} \right)}^{2}}{x} d x}=\int{\frac{4 \ln{\left(x \right)}^{2}}{x} d x}$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=4$$$ e $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}^{2}}{x}$$$:

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

Seja $$$u=\ln{\left(x \right)}$$$.

Então $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (veja os passos »), e obtemos $$$\frac{dx}{x} = du$$$.

Assim,

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

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=2$$$:

$$4 {\color{red}{\int{u^{2} d u}}}=4 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=4 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Recorde que $$$u=\ln{\left(x \right)}$$$:

$$\frac{4 {\color{red}{u}}^{3}}{3} = \frac{4 {\color{red}{\ln{\left(x \right)}}}^{3}}{3}$$

Portanto,

$$\int{\frac{4 \ln{\left(x \right)}^{2}}{x} d x} = \frac{4 \ln{\left(x \right)}^{3}}{3}$$

Adicione a constante de integração:

$$\int{\frac{4 \ln{\left(x \right)}^{2}}{x} d x} = \frac{4 \ln{\left(x \right)}^{3}}{3}+C$$

$$$\int \frac{\ln^{2}\left(x^{2}\right)}{x}\, dx = \frac{4 \ln^{3}\left(x\right)}{3} + C$$$A