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

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

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

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

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

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

A integral torna-se

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

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

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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