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

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

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

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

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

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

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

Thus,

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

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

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

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

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

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

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

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

Therefore,

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

Add the constant of integration:

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

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