Integral of $$$\frac{1}{x^{2} - 32 x}$$$

Find $$$\int \frac{1}{x^{2} - 32 x}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{x^{2} - 32 x} d x}}} = {\color{red}{\int{\left(\frac{1}{32 \left(x - 32\right)} - \frac{1}{32 x}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(\frac{1}{32 \left(x - 32\right)} - \frac{1}{32 x}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{32 x} d x} + \int{\frac{1}{32 \left(x - 32\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{32}$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$\int{\frac{1}{32 \left(x - 32\right)} d x} - {\color{red}{\int{\frac{1}{32 x} d x}}} = \int{\frac{1}{32 \left(x - 32\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{32}\right)}}$$

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

$$\int{\frac{1}{32 \left(x - 32\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{32} = \int{\frac{1}{32 \left(x - 32\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{32}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{32}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 32}$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{32} + {\color{red}{\int{\frac{1}{32 \left(x - 32\right)} d x}}} = - \frac{\ln{\left(\left|{x}\right| \right)}}{32} + {\color{red}{\left(\frac{\int{\frac{1}{x - 32} d x}}{32}\right)}}$$

Let $$$u=x - 32$$$.

Then $$$du=\left(x - 32\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{{\color{red}{\int{\frac{1}{x - 32} d x}}}}{32} = - \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{32}$$

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

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{32} = - \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{32}$$

Recall that $$$u=x - 32$$$:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{32} = - \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 32\right)}}}\right| \right)}}{32}$$

Therefore,

$$\int{\frac{1}{x^{2} - 32 x} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{32} + \frac{\ln{\left(\left|{x - 32}\right| \right)}}{32}$$

Simplify:

$$\int{\frac{1}{x^{2} - 32 x} d x} = \frac{- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 32}\right| \right)}}{32}$$

Add the constant of integration:

$$\int{\frac{1}{x^{2} - 32 x} d x} = \frac{- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 32}\right| \right)}}{32}+C$$

$$$\int \frac{1}{x^{2} - 32 x}\, dx = \frac{- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{x - 32}\right|\right)}{32} + C$$$A