Integraal van $$$\frac{8 x}{4 x^{2} - 5}$$$

Bepaal $$$\int \frac{8 x}{4 x^{2} - 5}\, dx$$$.

Zij $$$u=4 x^{2} - 5$$$.

Dan $$$du=\left(4 x^{2} - 5\right)^{\prime }dx = 8 x dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$x dx = \frac{du}{8}$$$.

Dus,

$${\color{red}{\int{\frac{8 x}{4 x^{2} - 5} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

We herinneren eraan dat $$$u=4 x^{2} - 5$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(4 x^{2} - 5\right)}}}\right| \right)}$$

Dus,

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

Voeg de integratieconstante toe:

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

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