Integraal van $$$\frac{4 x}{x^{2} - 1}$$$

Bepaal $$$\int \frac{4 x}{x^{2} - 1}\, dx$$$.

Zij $$$u=x^{2} - 1$$$.

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

De integraal wordt

$${\color{red}{\int{\frac{4 x}{x^{2} - 1} d x}}} = {\color{red}{\int{\frac{2}{u} d u}}}$$

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

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

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

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

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

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

Dus,

$$\int{\frac{4 x}{x^{2} - 1} d x} = 2 \ln{\left(\left|{x^{2} - 1}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{4 x}{x^{2} - 1} d x} = 2 \ln{\left(\left|{x^{2} - 1}\right| \right)}+C$$

$$$\int \frac{4 x}{x^{2} - 1}\, dx = 2 \ln\left(\left|{x^{2} - 1}\right|\right) + C$$$A