Integral of $$$\ln\left(u + v\right)$$$ with respect to $$$u$$$

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

Let $$$w=u + v$$$.

Then $$$dw=\left(u + v\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dw$$$.

The integral can be rewritten as

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

For the integral $$$\int{\ln{\left(w \right)} d w}$$$, use integration by parts $$$\int \operatorname{z} \operatorname{dl} = \operatorname{z}\operatorname{l} - \int \operatorname{l} \operatorname{dz}$$$.

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

Then $$$\operatorname{dz}=\left(\ln{\left(w \right)}\right)^{\prime }dw=\frac{dw}{w}$$$ (steps can be seen ») and $$$\operatorname{l}=\int{1 d w}=w$$$ (steps can be seen »).

The integral becomes

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

Apply the constant rule $$$\int c\, dw = c w$$$ with $$$c=1$$$:

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

Recall that $$$w=u + v$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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