INTEGRALES DE LA FORMA $\displaystyle \int R\left ( x,\sqrt[n]{\frac{ax+b}{cx+d}} \right )$

    Para resolver integrales de esta forma se utiliza la sustitución $\displaystyle y=\sqrt[m]{\frac{ax+b}{cx+d}}$, de donde se sigue que: $\displaystyle x=\frac{b-dy^{m}}{cy^{m}-a}$  y $\displaystyle dx=\frac{ad-bc}{(cy^{m}-a)^{2}}my^{m-1}dy$.

Ejemplo: 

     Calcular: $\displaystyle \int \sqrt{\frac{2+x}{2-x}}\frac{dx}{x}$ 

En este caso tomando: $\displaystyle y=\sqrt{\frac{2+x}{2-x}}$, se sigue que: $\displaystyle x=\frac{2(y^{2}-1)}{y^{2}+1}$  y $\displaystyle dx=8\frac{y}{(y^{2}+1)^{2}}dy$ y en consecuencia: 

$\displaystyle \int \sqrt{\frac{2+x}{2-x}}\frac{dx}{x}=\int y.\frac{8\frac{y}{(y^{2}+1)^{2}}dy}{\frac{2(y^{2}-1)}{y^{2}+1}}=4\int \frac{y^{2}}{(y^{2}+1)(y^{2}-1)}dy$ 

$\displaystyle =4\int \frac{y^{2}+1-1}{(y^{2}+1)(y^{2}-1)}dy=4\int \frac{1}{y^{2}-1}dy-4\int \frac{1}{(y^{2}+1)(y^{2}-1)}dy$ 

$\displaystyle =4\int\frac{1}{(y-1)(y+1)} dy-\int \frac{1}{(y-1)(y+1)(y^{2}+1)}dy$ 

$\displaystyle 2\int \frac{1}{y-1}dy-2\int \frac{dy}{y+1}-4\left [ \int \frac{\frac{1}{4}}{y-1}dy-\int \frac{\frac{1}{4}}{y+1}dy-\frac{1}{2}\int \frac{dy}{y^{2}+1} \right ]$ 

$\displaystyle 2ln\left | y-1 \right |-2ln\left | y+1 \right |-ln\left | y-1 \right |+ln\left | y+1 \right |+2arctan(y)+c$ 

$\displaystyle =ln\left | y-1 \right |-ln\left | y+1 \right |+2arctan(y)+c$ 

$\displaystyle =ln\left | \frac{y-1}{y+1} \right |+2arctan(y)+c$ 

$\displaystyle =ln\left | \frac{\sqrt{2+x}-\sqrt{2-x}}{\sqrt{2+x}+\sqrt{2-x}} \right | arctan\left ( \sqrt{\frac{2+x}{2-x}} \right )+ñc$