Integraal van $$$\frac{1}{x + 1}$$$

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

Zij $$$u=x + 1$$$.

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

De integraal wordt

$${\color{red}{\int{\frac{1}{x + 1} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

We herinneren eraan dat $$$u=x + 1$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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