Integraal van $$$\frac{1}{\left(x^{2} + 1\right)^{2}}$$$

Bepaal $$$\int \frac{1}{\left(x^{2} + 1\right)^{2}}\, dx$$$.

Om de integraal $$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}$$$ te berekenen, pas partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$ toe op de integraal $$$\int{\frac{1}{x^{2} + 1} d x}$$$.

Zij $$$\operatorname{u}=\frac{1}{x^{2} + 1}$$$ en $$$\operatorname{dv}=dx$$$.

Dan $$$\operatorname{du}=\left(\frac{1}{x^{2} + 1}\right)^{\prime }dx=- \frac{2 x}{\left(x^{2} + 1\right)^{2}} dx$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{1 d x}=x$$$ (de stappen zijn te zien »).

Dus,

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

Haal de constante eruit:

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

Herschrijf de teller van de integraand als $$$x^{2}=x^{2}{\color{red}{+1}}{\color{red}{-1}}$$$ en splits op:

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

Splits de integralen:

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

Zo krijgen we de volgende eenvoudige lineaire vergelijking met betrekking tot de integraal:

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

Door het op te lossen verkrijgen we dat

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

De integraal van $$$\frac{1}{x^{2} + 1}$$$ is $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

Dus,

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} = \frac{x}{2 \left(x^{2} + 1\right)} + \frac{\operatorname{atan}{\left(x \right)}}{2}$$

Vereenvoudig:

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} = \frac{x + \left(x^{2} + 1\right) \operatorname{atan}{\left(x \right)}}{2 \left(x^{2} + 1\right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} = \frac{x + \left(x^{2} + 1\right) \operatorname{atan}{\left(x \right)}}{2 \left(x^{2} + 1\right)}+C$$

$$$\int \frac{1}{\left(x^{2} + 1\right)^{2}}\, dx = \frac{x + \left(x^{2} + 1\right) \operatorname{atan}{\left(x \right)}}{2 \left(x^{2} + 1\right)} + C$$$A