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

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

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

Sean $$$\operatorname{u}=\ln{\left(x^{2} + 1 \right)}$$$ y $$$\operatorname{dv}=dx$$$.

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

Por lo tanto,

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

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

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

Reescribe y separa la fracción:

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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