Integraal van $$$\frac{x}{k - x^{2}}$$$ met betrekking tot $$$x$$$

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

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

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

De integraal kan worden herschreven als

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

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 $$$u=k - x^{2}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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