Integral of $$$\ln\left(x y\right)$$$ with respect to $$$x$$$

Find $$$\int \ln\left(x y\right)\, dx$$$.

Let $$$u=x y$$$.

Then $$$du=\left(x y\right)^{\prime }dx = y dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{y}$$$.

The integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{y}$$$ and $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

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

For the integral $$$\int{\ln{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{\omega} \operatorname{dv} = \operatorname{\omega}\operatorname{v} - \int \operatorname{v} \operatorname{d\omega}$$$.

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

Then $$$\operatorname{d\omega}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

Thus,

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Recall that $$$u=x y$$$:

$$\frac{- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{y} = \frac{- {\color{red}{x y}} + {\color{red}{x y}} \ln{\left({\color{red}{x y}} \right)}}{y}$$

Therefore,

$$\int{\ln{\left(x y \right)} d x} = \frac{x y \ln{\left(x y \right)} - x y}{y}$$

Simplify:

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

Add the constant of integration:

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

$$$\int \ln\left(x y\right)\, dx = x \left(\ln\left(x y\right) - 1\right) + C$$$A