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

Encontre $$$\int \ln\left(u\right)\, du$$$.

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

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

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

Portanto,

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

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

$$$\int \ln\left(u\right)\, du = u \left(\ln\left(u\right) - 1\right) + C$$$A