Integralen av $$$\frac{2 t}{\left(t - 3\right)^{2}}$$$

Bestäm $$$\int \frac{2 t}{\left(t - 3\right)^{2}}\, dt$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ med $$$c=2$$$ och $$$f{\left(t \right)} = \frac{t}{\left(t - 3\right)^{2}}$$$:

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

Skriv om integrandens täljare som $$$t=t - 3+3$$$ och dela upp bråket:

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

Integrera termvis:

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

Låt $$$u=t - 3$$$ vara.

Då $$$du=\left(t - 3\right)^{\prime }dt = 1 dt$$$ (stegen kan ses »), och vi har att $$$dt = du$$$.

Integralen blir

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

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Kom ihåg att $$$u=t - 3$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ med $$$c=3$$$ och $$$f{\left(t \right)} = \frac{1}{\left(t - 3\right)^{2}}$$$:

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

Låt $$$u=t - 3$$$ vara.

Då $$$du=\left(t - 3\right)^{\prime }dt = 1 dt$$$ (stegen kan ses »), och vi har att $$$dt = du$$$.

Integralen blir

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Kom ihåg att $$$u=t - 3$$$:

$$2 \ln{\left(\left|{t - 3}\right| \right)} - 6 {\color{red}{u}}^{-1} = 2 \ln{\left(\left|{t - 3}\right| \right)} - 6 {\color{red}{\left(t - 3\right)}}^{-1}$$

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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