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

Find $$$\int \frac{1}{- a + t}\, dt$$$.

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

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

Therefore,

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

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 $$$u=- a + t$$$:

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

Therefore,

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

Add the constant of integration:

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

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