Integral de $$$\ln\left(u + v\right)$$$ em relação a $$$u$$$

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

Seja $$$w=u + v$$$.

Então $$$dw=\left(u + v\right)^{\prime }du = 1 du$$$ (veja os passos »), e obtemos $$$du = dw$$$.

A integral torna-se

$${\color{red}{\int{\ln{\left(u + v \right)} d u}}} = {\color{red}{\int{\ln{\left(w \right)} d w}}}$$

Para a integral $$$\int{\ln{\left(w \right)} d w}$$$, use integração por partes $$$\int \operatorname{z} \operatorname{dl} = \operatorname{z}\operatorname{l} - \int \operatorname{l} \operatorname{dz}$$$.

Sejam $$$\operatorname{z}=\ln{\left(w \right)}$$$ e $$$\operatorname{dl}=dw$$$.

Então $$$\operatorname{dz}=\left(\ln{\left(w \right)}\right)^{\prime }dw=\frac{dw}{w}$$$ (os passos podem ser vistos ») e $$$\operatorname{l}=\int{1 d w}=w$$$ (os passos podem ser vistos »).

Assim,

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

Aplique a regra da constante $$$\int c\, dw = c w$$$ usando $$$c=1$$$:

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

Recorde que $$$w=u + v$$$:

$$- {\color{red}{w}} + {\color{red}{w}} \ln{\left({\color{red}{w}} \right)} = - {\color{red}{\left(u + v\right)}} + {\color{red}{\left(u + v\right)}} \ln{\left({\color{red}{\left(u + v\right)}} \right)}$$

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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