Integral of $$$\frac{1}{- a + b}$$$ with respect to $$$a$$$

Find $$$\int \frac{1}{- a + b}\, da$$$.

Let $$$u=- a + b$$$.

Then $$$du=\left(- a + b\right)^{\prime }da = - da$$$ (steps can be seen »), and we have that $$$da = - du$$$.

So,

$${\color{red}{\int{\frac{1}{- a + b} d a}}} = {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

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

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

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

Recall that $$$u=- a + b$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{{\color{red}{\left(- a + b\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{1}{- a + b} d a} = - \ln{\left(\left|{a - b}\right| \right)}$$

Add the constant of integration:

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

$$$\int \frac{1}{- a + b}\, da = - \ln\left(\left|{a - b}\right|\right) + C$$$A