Integral de $$$\frac{\ln\left(32 y\right)}{32 y}$$$

Encontre $$$\int \frac{\ln\left(32 y\right)}{32 y}\, dy$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ usando $$$c=\frac{1}{32}$$$ e $$$f{\left(y \right)} = \frac{\ln{\left(32 y \right)}}{y}$$$:

$${\color{red}{\int{\frac{\ln{\left(32 y \right)}}{32 y} d y}}} = {\color{red}{\left(\frac{\int{\frac{\ln{\left(32 y \right)}}{y} d y}}{32}\right)}}$$

Seja $$$u=\ln{\left(32 y \right)}$$$.

Então $$$du=\left(\ln{\left(32 y \right)}\right)^{\prime }dy = \frac{dy}{y}$$$ (veja os passos »), e obtemos $$$\frac{dy}{y} = du$$$.

A integral pode ser reescrita como

$$\frac{{\color{red}{\int{\frac{\ln{\left(32 y \right)}}{y} d y}}}}{32} = \frac{{\color{red}{\int{u d u}}}}{32}$$

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

$$\frac{{\color{red}{\int{u d u}}}}{32}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{32}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{32}$$

Recorde que $$$u=\ln{\left(32 y \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{64} = \frac{{\color{red}{\ln{\left(32 y \right)}}}^{2}}{64}$$

Portanto,

$$\int{\frac{\ln{\left(32 y \right)}}{32 y} d y} = \frac{\ln{\left(32 y \right)}^{2}}{64}$$

Simplifique:

$$\int{\frac{\ln{\left(32 y \right)}}{32 y} d y} = \frac{\left(\ln{\left(y \right)} + 5 \ln{\left(2 \right)}\right)^{2}}{64}$$

Adicione a constante de integração:

$$\int{\frac{\ln{\left(32 y \right)}}{32 y} d y} = \frac{\left(\ln{\left(y \right)} + 5 \ln{\left(2 \right)}\right)^{2}}{64}+C$$

$$$\int \frac{\ln\left(32 y\right)}{32 y}\, dy = \frac{\left(\ln\left(y\right) + 5 \ln\left(2\right)\right)^{2}}{64} + C$$$A