Integraal van $$$\frac{x^{3}}{m^{2} + 4 x^{2}}$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \frac{x^{3}}{m^{2} + 4 x^{2}}\, dx$$$.

Aangezien de graad van de teller niet kleiner is dan de graad van de noemer, voer een polynomiale staartdeling uit:

$${\color{red}{\int{\frac{x^{3}}{m^{2} + 4 x^{2}} d x}}} = {\color{red}{\int{\left(- \frac{m^{2} x}{4 \left(m^{2} + 4 x^{2}\right)} + \frac{x}{4}\right)d x}}}$$

Integreer termgewijs:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = x$$$:

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

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

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

Zij $$$u=4 m^{2} + 16 x^{2}$$$.

Dan $$$du=\left(4 m^{2} + 16 x^{2}\right)^{\prime }dx = 32 x dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$x dx = \frac{du}{32}$$$.

De integraal kan worden herschreven als

$$\frac{x^{2}}{8} - {\color{red}{\int{\frac{m^{2} x}{4 \left(m^{2} + 4 x^{2}\right)} d x}}} = \frac{x^{2}}{8} - {\color{red}{\int{\frac{m^{2}}{32 u} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{m^{2}}{32}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\frac{x^{2}}{8} - {\color{red}{\int{\frac{m^{2}}{32 u} d u}}} = \frac{x^{2}}{8} - {\color{red}{\left(\frac{m^{2} \int{\frac{1}{u} d u}}{32}\right)}}$$

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

$$- \frac{m^{2} {\color{red}{\int{\frac{1}{u} d u}}}}{32} + \frac{x^{2}}{8} = - \frac{m^{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{32} + \frac{x^{2}}{8}$$

We herinneren eraan dat $$$u=4 m^{2} + 16 x^{2}$$$:

$$- \frac{m^{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{32} + \frac{x^{2}}{8} = - \frac{m^{2} \ln{\left(\left|{{\color{red}{\left(4 m^{2} + 16 x^{2}\right)}}}\right| \right)}}{32} + \frac{x^{2}}{8}$$

Dus,

$$\int{\frac{x^{3}}{m^{2} + 4 x^{2}} d x} = - \frac{m^{2} \ln{\left(4 m^{2} + 16 x^{2} \right)}}{32} + \frac{x^{2}}{8}$$

Vereenvoudig:

$$\int{\frac{x^{3}}{m^{2} + 4 x^{2}} d x} = - \frac{m^{2} \left(\ln{\left(m^{2} + 4 x^{2} \right)} + 2 \ln{\left(2 \right)}\right)}{32} + \frac{x^{2}}{8}$$

Voeg de integratieconstante toe:

$$\int{\frac{x^{3}}{m^{2} + 4 x^{2}} d x} = - \frac{m^{2} \left(\ln{\left(m^{2} + 4 x^{2} \right)} + 2 \ln{\left(2 \right)}\right)}{32} + \frac{x^{2}}{8}+C$$

$$$\int \frac{x^{3}}{m^{2} + 4 x^{2}}\, dx = \left(- \frac{m^{2} \left(\ln\left(m^{2} + 4 x^{2}\right) + 2 \ln\left(2\right)\right)}{32} + \frac{x^{2}}{8}\right) + C$$$A