Integraal van $$$\frac{1}{1 - x} + \frac{1}{x \ln\left(x\right)}$$$

Bepaal $$$\int \left(\frac{1}{1 - x} + \frac{1}{x \ln\left(x\right)}\right)\, dx$$$.

Integreer termgewijs:

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

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

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

De integraal wordt

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

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

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

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

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

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

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

Zij $$$u=\ln{\left(x \right)}$$$.

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

De integraal wordt

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

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

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

We herinneren eraan dat $$$u=\ln{\left(x \right)}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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