Integral of $$$\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}$$$

Find $$$\int \frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}\, dx$$$.

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

$${\color{red}{\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x}}} = {\color{red}{\int{\left(\frac{x}{\left(x - 1\right)^{2}} - 2\right)d x}}}$$

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=2$$$:

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

Rewrite the numerator of the integrand as $$$x=x - 1+1$$$ and split the fraction:

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

Integrate term by term:

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

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

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

So,

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

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

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

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

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

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

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

Therefore,

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

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

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

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

Therefore,

$$\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x} = - 2 x + \ln{\left(\left|{x - 1}\right| \right)} - \frac{1}{x - 1}$$

Simplify:

$$\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x} = \frac{\left(- 2 x + \ln{\left(\left|{x - 1}\right| \right)}\right) \left(x - 1\right) - 1}{x - 1}$$

Add the constant of integration:

$$\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x} = \frac{\left(- 2 x + \ln{\left(\left|{x - 1}\right| \right)}\right) \left(x - 1\right) - 1}{x - 1}+C$$

$$$\int \frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}\, dx = \frac{\left(- 2 x + \ln\left(\left|{x - 1}\right|\right)\right) \left(x - 1\right) - 1}{x - 1} + C$$$A