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

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

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

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

Let $$$u=2 x - 1$$$.

Then $$$du=\left(2 x - 1\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

Therefore,

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

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

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

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 $$$u=2 x - 1$$$:

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

Therefore,

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

Add the constant of integration:

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

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