Integral of $$$\frac{1}{y \left(x - y\right)}$$$ with respect to $$$y$$$

Find $$$\int \frac{1}{y \left(x - y\right)}\, dy$$$.

Perform partial fraction decomposition:

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

Integrate term by term:

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

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

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

The integral of $$$\frac{1}{y}$$$ is $$$\int{\frac{1}{y} d y} = \ln{\left(\left|{y}\right| \right)}$$$:

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

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

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

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

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

Therefore,

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

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

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \frac{1}{y \left(x - y\right)}\, dy = \frac{\ln\left(\left|{y}\right|\right) - \ln\left(\left|{x - y}\right|\right)}{x} + C$$$A