Integral of $$$\frac{1}{15 - 3 x}$$$

Find $$$\int \frac{1}{15 - 3 x}\, dx$$$.

Let $$$u=15 - 3 x$$$.

Then $$$du=\left(15 - 3 x\right)^{\prime }dx = - 3 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{3}$$$.

So,

$${\color{red}{\int{\frac{1}{15 - 3 x} d x}}} = {\color{red}{\int{\left(- \frac{1}{3 u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{3}$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

$${\color{red}{\int{\left(- \frac{1}{3 u}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{u} d u}}{3}\right)}}$$

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

$$- \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{3} = - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{3}$$

Recall that $$$u=15 - 3 x$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{3} = - \frac{\ln{\left(\left|{{\color{red}{\left(15 - 3 x\right)}}}\right| \right)}}{3}$$

Therefore,

$$\int{\frac{1}{15 - 3 x} d x} = - \frac{\ln{\left(\left|{3 x - 15}\right| \right)}}{3}$$

Simplify:

$$\int{\frac{1}{15 - 3 x} d x} = - \frac{\ln{\left(\left|{x - 5}\right| \right)} + \ln{\left(3 \right)}}{3}$$

Add the constant of integration:

$$\int{\frac{1}{15 - 3 x} d x} = - \frac{\ln{\left(\left|{x - 5}\right| \right)} + \ln{\left(3 \right)}}{3}+C$$

$$$\int \frac{1}{15 - 3 x}\, dx = - \frac{\ln\left(\left|{x - 5}\right|\right) + \ln\left(3\right)}{3} + C$$$A