Integraal van $$$\frac{9}{5 - 4 x}$$$

Bepaal $$$\int \frac{9}{5 - 4 x}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=9$$$ en $$$f{\left(x \right)} = \frac{1}{5 - 4 x}$$$:

$${\color{red}{\int{\frac{9}{5 - 4 x} d x}}} = {\color{red}{\left(9 \int{\frac{1}{5 - 4 x} d x}\right)}}$$

Zij $$$u=5 - 4 x$$$.

Dan $$$du=\left(5 - 4 x\right)^{\prime }dx = - 4 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = - \frac{du}{4}$$$.

Dus,

$$9 {\color{red}{\int{\frac{1}{5 - 4 x} d x}}} = 9 {\color{red}{\int{\left(- \frac{1}{4 u}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{1}{4}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$9 {\color{red}{\int{\left(- \frac{1}{4 u}\right)d u}}} = 9 {\color{red}{\left(- \frac{\int{\frac{1}{u} d u}}{4}\right)}}$$

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

$$- \frac{9 {\color{red}{\int{\frac{1}{u} d u}}}}{4} = - \frac{9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

We herinneren eraan dat $$$u=5 - 4 x$$$:

$$- \frac{9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = - \frac{9 \ln{\left(\left|{{\color{red}{\left(5 - 4 x\right)}}}\right| \right)}}{4}$$

Dus,

$$\int{\frac{9}{5 - 4 x} d x} = - \frac{9 \ln{\left(\left|{4 x - 5}\right| \right)}}{4}$$

Voeg de integratieconstante toe:

$$\int{\frac{9}{5 - 4 x} d x} = - \frac{9 \ln{\left(\left|{4 x - 5}\right| \right)}}{4}+C$$

$$$\int \frac{9}{5 - 4 x}\, dx = - \frac{9 \ln\left(\left|{4 x - 5}\right|\right)}{4} + C$$$A