Funktion $$$-6 + \frac{1}{t^{3}}$$$ integraali

Määritä $$$\int \left(-6 + \frac{1}{t^{3}}\right)\, dt$$$.

Integroi termi kerrallaan:

$${\color{red}{\int{\left(-6 + \frac{1}{t^{3}}\right)d t}}} = {\color{red}{\left(- \int{6 d t} + \int{\frac{1}{t^{3}} d t}\right)}}$$

Sovella vakiosääntöä $$$\int c\, dt = c t$$$ käyttäen $$$c=6$$$:

$$\int{\frac{1}{t^{3}} d t} - {\color{red}{\int{6 d t}}} = \int{\frac{1}{t^{3}} d t} - {\color{red}{\left(6 t\right)}}$$

Sovella potenssisääntöä $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=-3$$$:

$$- 6 t + {\color{red}{\int{\frac{1}{t^{3}} d t}}}=- 6 t + {\color{red}{\int{t^{-3} d t}}}=- 6 t + {\color{red}{\frac{t^{-3 + 1}}{-3 + 1}}}=- 6 t + {\color{red}{\left(- \frac{t^{-2}}{2}\right)}}=- 6 t + {\color{red}{\left(- \frac{1}{2 t^{2}}\right)}}$$

Näin ollen,

$$\int{\left(-6 + \frac{1}{t^{3}}\right)d t} = - 6 t - \frac{1}{2 t^{2}}$$

Lisää integrointivakio:

$$\int{\left(-6 + \frac{1}{t^{3}}\right)d t} = - 6 t - \frac{1}{2 t^{2}}+C$$

$$$\int \left(-6 + \frac{1}{t^{3}}\right)\, dt = \left(- 6 t - \frac{1}{2 t^{2}}\right) + C$$$A