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

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

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

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

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,

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

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

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

$$$\int e^{2} \ln\left(x\right)\, dx = x \left(\ln\left(x\right) - 1\right) e^{2} + C$$$A