Integral of $$$\frac{2}{5 x - 1}$$$

Find $$$\int \frac{2}{5 x - 1}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \frac{1}{5 x - 1}$$$:

$${\color{red}{\int{\frac{2}{5 x - 1} d x}}} = {\color{red}{\left(2 \int{\frac{1}{5 x - 1} d x}\right)}}$$

Let $$$u=5 x - 1$$$.

Then $$$du=\left(5 x - 1\right)^{\prime }dx = 5 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{5}$$$.

Therefore,

$$2 {\color{red}{\int{\frac{1}{5 x - 1} d x}}} = 2 {\color{red}{\int{\frac{1}{5 u} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$2 {\color{red}{\int{\frac{1}{5 u} d u}}} = 2 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{5}\right)}}$$

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

$$\frac{2 {\color{red}{\int{\frac{1}{u} d u}}}}{5} = \frac{2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

Recall that $$$u=5 x - 1$$$:

$$\frac{2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} = \frac{2 \ln{\left(\left|{{\color{red}{\left(5 x - 1\right)}}}\right| \right)}}{5}$$

Therefore,

$$\int{\frac{2}{5 x - 1} d x} = \frac{2 \ln{\left(\left|{5 x - 1}\right| \right)}}{5}$$

Add the constant of integration:

$$\int{\frac{2}{5 x - 1} d x} = \frac{2 \ln{\left(\left|{5 x - 1}\right| \right)}}{5}+C$$

$$$\int \frac{2}{5 x - 1}\, dx = \frac{2 \ln\left(\left|{5 x - 1}\right|\right)}{5} + C$$$A