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

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

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=2$$$ dan $$$f{\left(x \right)} = \frac{x}{x - 1}$$$:

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

Tulis ulang dan pisahkan pecahannya:

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

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=1$$$:

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

Misalkan $$$u=x - 1$$$.

Kemudian $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = du$$$.

Integralnya menjadi

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

Integral dari $$$\frac{1}{u}$$$ adalah $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Ingat bahwa $$$u=x - 1$$$:

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

Oleh karena itu,

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

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

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