Integral de $$$\frac{1}{x^{2} \sqrt{x^{2} - 25}}$$$

Encontre $$$\int \frac{1}{x^{2} \sqrt{x^{2} - 25}}\, dx$$$.

Seja $$$x=5 \cosh{\left(u \right)}$$$.

Então $$$dx=\left(5 \cosh{\left(u \right)}\right)^{\prime }du = 5 \sinh{\left(u \right)} du$$$ (os passos podem ser vistos »).

Além disso, segue-se que $$$u=\operatorname{acosh}{\left(\frac{x}{5} \right)}$$$.

Logo,

$$$\frac{1}{x^{2} \sqrt{x^{2} - 25}} = \frac{1}{25 \sqrt{25 \cosh^{2}{\left( u \right)} - 25} \cosh^{2}{\left( u \right)}}$$$

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

$$$\frac{1}{25 \sqrt{25 \cosh^{2}{\left( u \right)} - 25} \cosh^{2}{\left( u \right)}}=\frac{1}{125 \sqrt{\cosh^{2}{\left( u \right)} - 1} \cosh^{2}{\left( u \right)}}=\frac{1}{125 \sqrt{\sinh^{2}{\left( u \right)}} \cosh^{2}{\left( u \right)}}$$$

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

$$$\frac{1}{125 \sqrt{\sinh^{2}{\left( u \right)}} \cosh^{2}{\left( u \right)}} = \frac{1}{125 \sinh{\left( u \right)} \cosh^{2}{\left( u \right)}}$$$

Portanto,

$${\color{red}{\int{\frac{1}{x^{2} \sqrt{x^{2} - 25}} d x}}} = {\color{red}{\int{\frac{1}{25 \cosh^{2}{\left(u \right)}} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{25}$$$ e $$$f{\left(u \right)} = \frac{1}{\cosh^{2}{\left(u \right)}}$$$:

$${\color{red}{\int{\frac{1}{25 \cosh^{2}{\left(u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}{25}\right)}}$$

Reescreva o integrando em termos da secante hiperbólica.:

$$\frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}}}{25} = \frac{{\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{25}$$

A integral de $$$\operatorname{sech}^{2}{\left(u \right)}$$$ é $$$\int{\operatorname{sech}^{2}{\left(u \right)} d u} = \tanh{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{25} = \frac{{\color{red}{\tanh{\left(u \right)}}}}{25}$$

Recorde que $$$u=\operatorname{acosh}{\left(\frac{x}{5} \right)}$$$:

$$\frac{\tanh{\left({\color{red}{u}} \right)}}{25} = \frac{\tanh{\left({\color{red}{\operatorname{acosh}{\left(\frac{x}{5} \right)}}} \right)}}{25}$$

Portanto,

$$\int{\frac{1}{x^{2} \sqrt{x^{2} - 25}} d x} = \frac{\sqrt{\frac{x}{5} - 1} \sqrt{\frac{x}{5} + 1}}{5 x}$$

Simplifique:

$$\int{\frac{1}{x^{2} \sqrt{x^{2} - 25}} d x} = \frac{\sqrt{x - 5} \sqrt{x + 5}}{25 x}$$

Adicione a constante de integração:

$$\int{\frac{1}{x^{2} \sqrt{x^{2} - 25}} d x} = \frac{\sqrt{x - 5} \sqrt{x + 5}}{25 x}+C$$

$$$\int \frac{1}{x^{2} \sqrt{x^{2} - 25}}\, dx = \frac{\sqrt{x - 5} \sqrt{x + 5}}{25 x} + C$$$A