Integraal van $$$\frac{x^{3}}{2 \left(25 - x^{2}\right)}$$$

Bepaal $$$\int \frac{x^{3}}{2 \left(25 - x^{2}\right)}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(x \right)} = \frac{x^{3}}{25 - x^{2}}$$$:

$${\color{red}{\int{\frac{x^{3}}{2 \left(25 - x^{2}\right)} d x}}} = {\color{red}{\left(\frac{\int{\frac{x^{3}}{25 - x^{2}} d x}}{2}\right)}}$$

Aangezien de graad van de teller niet kleiner is dan die van de noemer, voer een staartdeling van polynomen uit (stappen zijn te zien »):

$$\frac{{\color{red}{\int{\frac{x^{3}}{25 - x^{2}} d x}}}}{2} = \frac{{\color{red}{\int{\left(- x + \frac{25 x}{25 - x^{2}}\right)d x}}}}{2}$$

Integreer termgewijs:

$$\frac{{\color{red}{\int{\left(- x + \frac{25 x}{25 - x^{2}}\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{x d x} + \int{\frac{25 x}{25 - x^{2}} d x}\right)}}}{2}$$

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

$$\frac{\int{\frac{25 x}{25 - x^{2}} d x}}{2} - \frac{{\color{red}{\int{x d x}}}}{2}=\frac{\int{\frac{25 x}{25 - x^{2}} d x}}{2} - \frac{{\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{2}=\frac{\int{\frac{25 x}{25 - x^{2}} d x}}{2} - \frac{{\color{red}{\left(\frac{x^{2}}{2}\right)}}}{2}$$

Zij $$$u=25 - x^{2}$$$.

Dan $$$du=\left(25 - x^{2}\right)^{\prime }dx = - 2 x dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$x dx = - \frac{du}{2}$$$.

Dus,

$$- \frac{x^{2}}{4} + \frac{{\color{red}{\int{\frac{25 x}{25 - x^{2}} d x}}}}{2} = - \frac{x^{2}}{4} + \frac{{\color{red}{\int{\left(- \frac{25}{2 u}\right)d u}}}}{2}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{25}{2}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$- \frac{x^{2}}{4} + \frac{{\color{red}{\int{\left(- \frac{25}{2 u}\right)d u}}}}{2} = - \frac{x^{2}}{4} + \frac{{\color{red}{\left(- \frac{25 \int{\frac{1}{u} d u}}{2}\right)}}}{2}$$

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

$$- \frac{x^{2}}{4} - \frac{25 {\color{red}{\int{\frac{1}{u} d u}}}}{4} = - \frac{x^{2}}{4} - \frac{25 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

We herinneren eraan dat $$$u=25 - x^{2}$$$:

$$- \frac{x^{2}}{4} - \frac{25 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = - \frac{x^{2}}{4} - \frac{25 \ln{\left(\left|{{\color{red}{\left(25 - x^{2}\right)}}}\right| \right)}}{4}$$

Dus,

$$\int{\frac{x^{3}}{2 \left(25 - x^{2}\right)} d x} = - \frac{x^{2}}{4} - \frac{25 \ln{\left(\left|{x^{2} - 25}\right| \right)}}{4}$$

Voeg de integratieconstante toe:

$$\int{\frac{x^{3}}{2 \left(25 - x^{2}\right)} d x} = - \frac{x^{2}}{4} - \frac{25 \ln{\left(\left|{x^{2} - 25}\right| \right)}}{4}+C$$

$$$\int \frac{x^{3}}{2 \left(25 - x^{2}\right)}\, dx = \left(- \frac{x^{2}}{4} - \frac{25 \ln\left(\left|{x^{2} - 25}\right|\right)}{4}\right) + C$$$A