Integral de $$$\ln\left(y\right)$$$

Encontre $$$\int \ln\left(y\right)\, dy$$$.

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

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

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

Assim,

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

Aplique a regra da constante $$$\int c\, dy = c y$$$ usando $$$c=1$$$:

$$y \ln{\left(y \right)} - {\color{red}{\int{1 d y}}} = y \ln{\left(y \right)} - {\color{red}{y}}$$

Portanto,

$$\int{\ln{\left(y \right)} d y} = y \ln{\left(y \right)} - y$$

Simplifique:

$$\int{\ln{\left(y \right)} d y} = y \left(\ln{\left(y \right)} - 1\right)$$

Adicione a constante de integração:

$$\int{\ln{\left(y \right)} d y} = y \left(\ln{\left(y \right)} - 1\right)+C$$

$$$\int \ln\left(y\right)\, dy = y \left(\ln\left(y\right) - 1\right) + C$$$A