Funktion $$$\frac{2500 - 3 t}{t^{2}}$$$ integraali

Määritä $$$\int \frac{2500 - 3 t}{t^{2}}\, dt$$$.

Expand the expression:

$${\color{red}{\int{\frac{2500 - 3 t}{t^{2}} d t}}} = {\color{red}{\int{\left(- \frac{3}{t} + \frac{2500}{t^{2}}\right)d t}}}$$

Integroi termi kerrallaan:

$${\color{red}{\int{\left(- \frac{3}{t} + \frac{2500}{t^{2}}\right)d t}}} = {\color{red}{\left(\int{\frac{2500}{t^{2}} d t} - \int{\frac{3}{t} d t}\right)}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ käyttäen $$$c=3$$$ ja $$$f{\left(t \right)} = \frac{1}{t}$$$:

$$\int{\frac{2500}{t^{2}} d t} - {\color{red}{\int{\frac{3}{t} d t}}} = \int{\frac{2500}{t^{2}} d t} - {\color{red}{\left(3 \int{\frac{1}{t} d t}\right)}}$$

Funktion $$$\frac{1}{t}$$$ integraali on $$$\int{\frac{1}{t} d t} = \ln{\left(\left|{t}\right| \right)}$$$:

$$\int{\frac{2500}{t^{2}} d t} - 3 {\color{red}{\int{\frac{1}{t} d t}}} = \int{\frac{2500}{t^{2}} d t} - 3 {\color{red}{\ln{\left(\left|{t}\right| \right)}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ käyttäen $$$c=2500$$$ ja $$$f{\left(t \right)} = \frac{1}{t^{2}}$$$:

$$- 3 \ln{\left(\left|{t}\right| \right)} + {\color{red}{\int{\frac{2500}{t^{2}} d t}}} = - 3 \ln{\left(\left|{t}\right| \right)} + {\color{red}{\left(2500 \int{\frac{1}{t^{2}} d t}\right)}}$$

Sovella potenssisääntöä $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=-2$$$:

$$- 3 \ln{\left(\left|{t}\right| \right)} + 2500 {\color{red}{\int{\frac{1}{t^{2}} d t}}}=- 3 \ln{\left(\left|{t}\right| \right)} + 2500 {\color{red}{\int{t^{-2} d t}}}=- 3 \ln{\left(\left|{t}\right| \right)} + 2500 {\color{red}{\frac{t^{-2 + 1}}{-2 + 1}}}=- 3 \ln{\left(\left|{t}\right| \right)} + 2500 {\color{red}{\left(- t^{-1}\right)}}=- 3 \ln{\left(\left|{t}\right| \right)} + 2500 {\color{red}{\left(- \frac{1}{t}\right)}}$$

Näin ollen,

$$\int{\frac{2500 - 3 t}{t^{2}} d t} = - 3 \ln{\left(\left|{t}\right| \right)} - \frac{2500}{t}$$

Lisää integrointivakio:

$$\int{\frac{2500 - 3 t}{t^{2}} d t} = - 3 \ln{\left(\left|{t}\right| \right)} - \frac{2500}{t}+C$$

$$$\int \frac{2500 - 3 t}{t^{2}}\, dt = \left(- 3 \ln\left(\left|{t}\right|\right) - \frac{2500}{t}\right) + C$$$A