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

Find $$$\int \frac{1}{t^{2} - 2 t}\, dt$$$.

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

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

Integrate term by term:

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

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

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

Let $$$u=t - 2$$$.

Then $$$du=\left(t - 2\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

Thus,

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

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

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

Recall that $$$u=t - 2$$$:

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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