Integral de $$$\ln\left(u + v\right)$$$ con respecto a $$$u$$$

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

Sea $$$w=u + v$$$.

Entonces $$$dw=\left(u + v\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dw$$$.

Por lo tanto,

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

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

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

Entonces $$$\operatorname{dz}=\left(\ln{\left(w \right)}\right)^{\prime }dw=\frac{dw}{w}$$$ (los pasos pueden verse ») y $$$\operatorname{l}=\int{1 d w}=w$$$ (los pasos pueden verse »).

La integral se convierte en

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

Aplica la regla de la constante $$$\int c\, dw = c w$$$ con $$$c=1$$$:

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

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

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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