Integral of $$$\frac{1}{15 - t}$$$

Find $$$\int \frac{1}{15 - t}\, dt$$$.

Let $$$u=15 - t$$$.

Then $$$du=\left(15 - t\right)^{\prime }dt = - dt$$$ (steps can be seen »), and we have that $$$dt = - du$$$.

Thus,

$${\color{red}{\int{\frac{1}{15 - t} d t}}} = {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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 $$$u=15 - t$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{{\color{red}{\left(15 - t\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{1}{15 - t} d t} = - \ln{\left(\left|{t - 15}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{1}{15 - t} d t} = - \ln{\left(\left|{t - 15}\right| \right)}+C$$

$$$\int \frac{1}{15 - t}\, dt = - \ln\left(\left|{t - 15}\right|\right) + C$$$A