Integraal van $$$\frac{x}{\left(x - 1\right)^{2}}$$$

Bepaal $$$\int \frac{x}{\left(x - 1\right)^{2}}\, dx$$$.

Herschrijf de teller van de integraand als $$$x=x - 1+1$$$ en splits de breuk:

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

Integreer termgewijs:

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

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

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

Dus,

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

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

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

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

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

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 kan worden herschreven als

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

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

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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