Integraal van $$$\frac{1}{1 - 2 x}$$$

Bepaal $$$\int \frac{1}{1 - 2 x}\, dx$$$.

Zij $$$u=1 - 2 x$$$.

Dan $$$du=\left(1 - 2 x\right)^{\prime }dx = - 2 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = - \frac{du}{2}$$$.

Dus,

$${\color{red}{\int{\frac{1}{1 - 2 x} d x}}} = {\color{red}{\int{\left(- \frac{1}{2 u}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{1}{2}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

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

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

We herinneren eraan dat $$$u=1 - 2 x$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = - \frac{\ln{\left(\left|{{\color{red}{\left(1 - 2 x\right)}}}\right| \right)}}{2}$$

Dus,

$$\int{\frac{1}{1 - 2 x} d x} = - \frac{\ln{\left(\left|{2 x - 1}\right| \right)}}{2}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{1 - 2 x} d x} = - \frac{\ln{\left(\left|{2 x - 1}\right| \right)}}{2}+C$$

$$$\int \frac{1}{1 - 2 x}\, dx = - \frac{\ln\left(\left|{2 x - 1}\right|\right)}{2} + C$$$A