Integral de $$$y \ln\left(y\right) + 1$$$

Encontre $$$\int \left(y \ln\left(y\right) + 1\right)\, dy$$$.

Integre termo a termo:

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

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

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

Para a integral $$$\int{y \ln{\left(y \right)} d y}$$$, 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 $$$\operatorname{dv}=y dy$$$.

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

Logo,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(y \right)} = y$$$:

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

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

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

Portanto,

$$\int{\left(y \ln{\left(y \right)} + 1\right)d y} = \frac{y^{2} \ln{\left(y \right)}}{2} - \frac{y^{2}}{4} + y$$

Simplifique:

$$\int{\left(y \ln{\left(y \right)} + 1\right)d y} = \frac{y \left(2 y \ln{\left(y \right)} - y + 4\right)}{4}$$

Adicione a constante de integração:

$$\int{\left(y \ln{\left(y \right)} + 1\right)d y} = \frac{y \left(2 y \ln{\left(y \right)} - y + 4\right)}{4}+C$$

$$$\int \left(y \ln\left(y\right) + 1\right)\, dy = \frac{y \left(2 y \ln\left(y\right) - y + 4\right)}{4} + C$$$A