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

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

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

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

Integrate term by term:

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

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

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

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

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

The integral becomes

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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