Integral of $$$\frac{t^{3} - 1}{t}$$$

Find $$$\int \frac{t^{3} - 1}{t}\, dt$$$.

Expand the expression:

$${\color{red}{\int{\frac{t^{3} - 1}{t} d t}}} = {\color{red}{\int{\left(t^{2} - \frac{1}{t}\right)d t}}}$$

Integrate term by term:

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

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

$$- \int{\frac{1}{t} d t} + {\color{red}{\int{t^{2} d t}}}=- \int{\frac{1}{t} d t} + {\color{red}{\frac{t^{1 + 2}}{1 + 2}}}=- \int{\frac{1}{t} d t} + {\color{red}{\left(\frac{t^{3}}{3}\right)}}$$

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

$$\frac{t^{3}}{3} - {\color{red}{\int{\frac{1}{t} d t}}} = \frac{t^{3}}{3} - {\color{red}{\ln{\left(\left|{t}\right| \right)}}}$$

Therefore,

$$\int{\frac{t^{3} - 1}{t} d t} = \frac{t^{3}}{3} - \ln{\left(\left|{t}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{t^{3} - 1}{t} d t} = \frac{t^{3}}{3} - \ln{\left(\left|{t}\right| \right)}+C$$

$$$\int \frac{t^{3} - 1}{t}\, dt = \left(\frac{t^{3}}{3} - \ln\left(\left|{t}\right|\right)\right) + C$$$A