Integral de $$$2 \operatorname{atan}{\left(x \right)}$$$

Halla $$$\int 2 \operatorname{atan}{\left(x \right)}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=2$$$ y $$$f{\left(x \right)} = \operatorname{atan}{\left(x \right)}$$$:

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

Para la integral $$$\int{\operatorname{atan}{\left(x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\operatorname{atan}{\left(x \right)}$$$ y $$$\operatorname{dv}=dx$$$.

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

Entonces,

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

Sea $$$u=x^{2} + 1$$$.

Entonces $$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (los pasos pueden verse »), y obtenemos que $$$x dx = \frac{du}{2}$$$.

La integral se convierte en

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$2 x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = 2 x \operatorname{atan}{\left(x \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recordemos que $$$u=x^{2} + 1$$$:

$$2 x \operatorname{atan}{\left(x \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 2 x \operatorname{atan}{\left(x \right)} - \ln{\left(\left|{{\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}$$

Por lo tanto,

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

Añade la constante de integración:

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

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