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

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

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

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

Karena derajat pembilang tidak kurang dari derajat penyebut, lakukan pembagian panjang polinom (langkah-langkah dapat dilihat »):

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

Integralkan suku demi suku:

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

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

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

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

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

Integral tersebut dapat ditulis ulang sebagai

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

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

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

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

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

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

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

Terapkan aturan pangkat $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=1$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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