$$$\frac{t^{3} - 1}{t}$$$の積分

$$$\int \frac{t^{3} - 1}{t}\, dt$$$ を求めよ。

Expand the expression:

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

項別に積分せよ:

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

$$$n=2$$$ を用いて、べき乗の法則 $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

$$$\frac{1}{t}$$$ の不定積分は $$$\int{\frac{1}{t} d t} = \ln{\left(\left|{t}\right| \right)}$$$ です:

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

したがって、

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

積分定数を加える:

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

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