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

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

Let $$$u=a - p$$$.

Then $$$du=\left(a - p\right)^{\prime }da = 1 da$$$ (steps can be seen »), and we have that $$$da = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{1}{a - p} d a}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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 $$$u=a - p$$$:

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

Therefore,

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

Add the constant of integration:

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

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