Integral de $$$\frac{1}{\left(x^{2} + 1\right)^{2}}$$$

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

Para calcular la integral $$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}$$$, aplique la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$ a la integral $$$\int{\frac{1}{x^{2} + 1} d x}$$$.

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

Entonces $$$\operatorname{du}=\left(\frac{1}{x^{2} + 1}\right)^{\prime }dx=- \frac{2 x}{\left(x^{2} + 1\right)^{2}} dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d x}=x$$$ (los pasos pueden verse »).

Por lo tanto,

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

Extrae la constante:

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

Reescribe el numerador del integrando como $$$x^{2}=x^{2}{\color{red}{+1}}{\color{red}{-1}}$$$ y descompón:

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

Separa las integrales:

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

Por lo tanto, obtenemos la siguiente ecuación lineal simple con respecto a la integral:

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

Al resolverlo, obtenemos que

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

La integral de $$$\frac{1}{x^{2} + 1}$$$ es $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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