Integral de $$$\frac{1}{\sqrt{4 x^{2} - 5}}$$$

Halla $$$\int \frac{1}{\sqrt{4 x^{2} - 5}}\, dx$$$.

Sea $$$x=\frac{\sqrt{5} \cosh{\left(u \right)}}{2}$$$.

Entonces $$$dx=\left(\frac{\sqrt{5} \cosh{\left(u \right)}}{2}\right)^{\prime }du = \frac{\sqrt{5} \sinh{\left(u \right)}}{2} du$$$ (los pasos pueden verse »).

Además, se sigue que $$$u=\operatorname{acosh}{\left(\frac{2 \sqrt{5} x}{5} \right)}$$$.

Por lo tanto,

$$$\frac{1}{\sqrt{4 x^{2} - 5}} = \frac{1}{\sqrt{5 \cosh^{2}{\left( u \right)} - 5}}$$$

Utiliza la identidad $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{5 \cosh^{2}{\left( u \right)} - 5}}=\frac{\sqrt{5}}{5 \sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{\sqrt{5}}{5 \sqrt{\sinh^{2}{\left( u \right)}}}$$$

Suponiendo que $$$\sinh{\left( u \right)} \ge 0$$$, obtenemos lo siguiente:

$$$\frac{\sqrt{5}}{5 \sqrt{\sinh^{2}{\left( u \right)}}} = \frac{\sqrt{5}}{5 \sinh{\left( u \right)}}$$$

Entonces,

$${\color{red}{\int{\frac{1}{\sqrt{4 x^{2} - 5}} d x}}} = {\color{red}{\int{\frac{1}{2} d u}}}$$

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=\frac{1}{2}$$$:

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

Recordemos que $$$u=\operatorname{acosh}{\left(\frac{2 \sqrt{5} x}{5} \right)}$$$:

$$\frac{{\color{red}{u}}}{2} = \frac{{\color{red}{\operatorname{acosh}{\left(\frac{2 \sqrt{5} x}{5} \right)}}}}{2}$$

Por lo tanto,

$$\int{\frac{1}{\sqrt{4 x^{2} - 5}} d x} = \frac{\operatorname{acosh}{\left(\frac{2 \sqrt{5} x}{5} \right)}}{2}$$

Añade la constante de integración:

$$\int{\frac{1}{\sqrt{4 x^{2} - 5}} d x} = \frac{\operatorname{acosh}{\left(\frac{2 \sqrt{5} x}{5} \right)}}{2}+C$$

$$$\int \frac{1}{\sqrt{4 x^{2} - 5}}\, dx = \frac{\operatorname{acosh}{\left(\frac{2 \sqrt{5} x}{5} \right)}}{2} + C$$$A