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

Find $$$\int \frac{2 v}{v - 1}\, dv$$$.

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

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:

$$2 \int{\frac{1}{v - 1} d v} + 2 {\color{red}{\int{1 d v}}} = 2 \int{\frac{1}{v - 1} d v} + 2 {\color{red}{v}}$$

Let $$$u=v - 1$$$.

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

The integral becomes

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

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

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

Recall that $$$u=v - 1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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