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

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

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=2$$$ e $$$f{\left(v \right)} = \frac{v}{v - 1}$$$:

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

Reescreva e separe a fração:

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dv = c v$$$ usando $$$c=1$$$:

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

Seja $$$u=v - 1$$$.

Então $$$du=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (veja os passos »), e obtemos $$$dv = du$$$.

Assim,

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

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recorde que $$$u=v - 1$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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