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

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

Perform partial fraction decomposition (steps can be seen »):

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

Integrate term by term:

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

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

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

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

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

Thus,

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

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

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

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

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

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

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

Thus,

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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