Integraal van $$$\frac{2 t - 7}{t - 8}$$$

Bepaal $$$\int \frac{2 t - 7}{t - 8}\, dt$$$.

Herschrijf de teller van de integraand als $$$2 t - 7=2\left(t - 8\right)+9$$$ en splits de breuk:

$${\color{red}{\int{\frac{2 t - 7}{t - 8} d t}}} = {\color{red}{\int{\left(2 + \frac{9}{t - 8}\right)d t}}}$$

Integreer termgewijs:

$${\color{red}{\int{\left(2 + \frac{9}{t - 8}\right)d t}}} = {\color{red}{\left(\int{2 d t} + \int{\frac{9}{t - 8} d t}\right)}}$$

Pas de constantenregel $$$\int c\, dt = c t$$$ toe met $$$c=2$$$:

$$\int{\frac{9}{t - 8} d t} + {\color{red}{\int{2 d t}}} = \int{\frac{9}{t - 8} d t} + {\color{red}{\left(2 t\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=9$$$ en $$$f{\left(t \right)} = \frac{1}{t - 8}$$$:

$$2 t + {\color{red}{\int{\frac{9}{t - 8} d t}}} = 2 t + {\color{red}{\left(9 \int{\frac{1}{t - 8} d t}\right)}}$$

Zij $$$u=t - 8$$$.

Dan $$$du=\left(t - 8\right)^{\prime }dt = 1 dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$dt = du$$$.

De integraal wordt

$$2 t + 9 {\color{red}{\int{\frac{1}{t - 8} d t}}} = 2 t + 9 {\color{red}{\int{\frac{1}{u} d u}}}$$

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$2 t + 9 {\color{red}{\int{\frac{1}{u} d u}}} = 2 t + 9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

We herinneren eraan dat $$$u=t - 8$$$:

$$2 t + 9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 2 t + 9 \ln{\left(\left|{{\color{red}{\left(t - 8\right)}}}\right| \right)}$$

Dus,

$$\int{\frac{2 t - 7}{t - 8} d t} = 2 t + 9 \ln{\left(\left|{t - 8}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{2 t - 7}{t - 8} d t} = 2 t + 9 \ln{\left(\left|{t - 8}\right| \right)}+C$$

$$$\int \frac{2 t - 7}{t - 8}\, dt = \left(2 t + 9 \ln\left(\left|{t - 8}\right|\right)\right) + C$$$A