Integraal van $$$\frac{5 v}{1 - 4 v^{2}}$$$

Bepaal $$$\int \frac{5 v}{1 - 4 v^{2}}\, dv$$$.

Zij $$$u=1 - 4 v^{2}$$$.

Dan $$$du=\left(1 - 4 v^{2}\right)^{\prime }dv = - 8 v dv$$$ (de stappen zijn te zien »), en dan geldt dat $$$v dv = - \frac{du}{8}$$$.

Dus,

$${\color{red}{\int{\frac{5 v}{1 - 4 v^{2}} d v}}} = {\color{red}{\int{\left(- \frac{5}{8 u}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{5}{8}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$${\color{red}{\int{\left(- \frac{5}{8 u}\right)d u}}} = {\color{red}{\left(- \frac{5 \int{\frac{1}{u} d u}}{8}\right)}}$$

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

$$- \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = - \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

We herinneren eraan dat $$$u=1 - 4 v^{2}$$$:

$$- \frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = - \frac{5 \ln{\left(\left|{{\color{red}{\left(1 - 4 v^{2}\right)}}}\right| \right)}}{8}$$

Dus,

$$\int{\frac{5 v}{1 - 4 v^{2}} d v} = - \frac{5 \ln{\left(\left|{4 v^{2} - 1}\right| \right)}}{8}$$

Voeg de integratieconstante toe:

$$\int{\frac{5 v}{1 - 4 v^{2}} d v} = - \frac{5 \ln{\left(\left|{4 v^{2} - 1}\right| \right)}}{8}+C$$

$$$\int \frac{5 v}{1 - 4 v^{2}}\, dv = - \frac{5 \ln\left(\left|{4 v^{2} - 1}\right|\right)}{8} + C$$$A