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

Encontre $$$\int \operatorname{atanh}{\left(x \right)}\, dx$$$.

Para a integral $$$\int{\operatorname{atanh}{\left(x \right)} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\operatorname{atanh}{\left(x \right)}$$$ e $$$\operatorname{dv}=dx$$$.

Então $$$\operatorname{du}=\left(\operatorname{atanh}{\left(x \right)}\right)^{\prime }dx=- \frac{1}{x^{2} - 1} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{1 d x}=x$$$ (os passos podem ser vistos »).

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-1$$$ e $$$f{\left(x \right)} = \frac{x}{\left(x - 1\right) \left(x + 1\right)}$$$:

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

Efetue a decomposição em frações parciais (os passos podem ser vistos »):

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

Integre termo a termo:

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(x \right)} = \frac{1}{x + 1}$$$:

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

Seja $$$u=x + 1$$$.

Então $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Logo,

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

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

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

Recorde que $$$u=x + 1$$$:

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

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

Seja $$$u=x - 1$$$.

Então $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

A integral pode ser reescrita como

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

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

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

Recorde que $$$u=x - 1$$$:

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

Portanto,

$$\int{\operatorname{atanh}{\left(x \right)} d x} = x \operatorname{atanh}{\left(x \right)} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}$$

Adicione a constante de integração:

$$\int{\operatorname{atanh}{\left(x \right)} d x} = x \operatorname{atanh}{\left(x \right)} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}+C$$

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