Integral de $$$\frac{\sqrt{x^{2} - 1}}{x^{2}}$$$

Encontre $$$\int \frac{\sqrt{x^{2} - 1}}{x^{2}}\, dx$$$.

Seja $$$x=\cosh{\left(u \right)}$$$.

Então $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (os passos podem ser vistos »).

Além disso, segue-se que $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Logo,

$$$\frac{\sqrt{x^{2} - 1}}{x^{2}} = \frac{\sqrt{\cosh^{2}{\left( u \right)} - 1}}{\cosh^{2}{\left( u \right)}}$$$

Use a identidade $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

$$$\frac{\sqrt{\cosh^{2}{\left( u \right)} - 1}}{\cosh^{2}{\left( u \right)}}=\frac{\sqrt{\sinh^{2}{\left( u \right)}}}{\cosh^{2}{\left( u \right)}}$$$

Supondo que $$$\sinh{\left( u \right)} \ge 0$$$, obtemos o seguinte:

$$$\frac{\sqrt{\sinh^{2}{\left( u \right)}}}{\cosh^{2}{\left( u \right)}} = \frac{\sinh{\left( u \right)}}{\cosh^{2}{\left( u \right)}}$$$

Logo,

$${\color{red}{\int{\frac{\sqrt{x^{2} - 1}}{x^{2}} d x}}} = {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)}}{\cosh^{2}{\left(u \right)}} d u}}}$$

Reescreva em termos da tangente hiperbólica:

$${\color{red}{\int{\frac{\sinh^{2}{\left(u \right)}}{\cosh^{2}{\left(u \right)}} d u}}} = {\color{red}{\int{\tanh^{2}{\left(u \right)} d u}}}$$

Seja $$$v=\tanh{\left(u \right)}$$$.

Então $$$dv=\left(\tanh{\left(u \right)}\right)^{\prime }du = \operatorname{sech}^{2}{\left(u \right)} du$$$ (veja os passos »), e obtemos $$$\operatorname{sech}^{2}{\left(u \right)} du = dv$$$.

Portanto,

$${\color{red}{\int{\tanh^{2}{\left(u \right)} d u}}} = {\color{red}{\int{\left(- \frac{v^{2}}{v^{2} - 1}\right)d v}}}$$

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

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

Reescreva e separe a fração:

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dv = c v$$$ usando $$$c=1$$$:

$$- \int{\frac{1}{v^{2} - 1} d v} - {\color{red}{\int{1 d v}}} = - \int{\frac{1}{v^{2} - 1} d v} - {\color{red}{v}}$$

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

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

Integre termo a termo:

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

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

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

Seja $$$w=v - 1$$$.

Então $$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (veja os passos »), e obtemos $$$dv = dw$$$.

Portanto,

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

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

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

Recorde que $$$w=v - 1$$$:

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

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

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

Seja $$$w=v + 1$$$.

Então $$$dw=\left(v + 1\right)^{\prime }dv = 1 dv$$$ (veja os passos »), e obtemos $$$dv = dw$$$.

Logo,

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

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

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

Recorde que $$$w=v + 1$$$:

$$- v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} = - v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(v + 1\right)}}}\right| \right)}}{2}$$

Recorde que $$$v=\tanh{\left(u \right)}$$$:

$$- \frac{\ln{\left(\left|{-1 + {\color{red}{v}}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + {\color{red}{v}}}\right| \right)}}{2} - {\color{red}{v}} = - \frac{\ln{\left(\left|{-1 + {\color{red}{\tanh{\left(u \right)}}}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + {\color{red}{\tanh{\left(u \right)}}}}\right| \right)}}{2} - {\color{red}{\tanh{\left(u \right)}}}$$

Recorde que $$$u=\operatorname{acosh}{\left(x \right)}$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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