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

Halla $$$\int \frac{\sqrt{1 - x}}{x}\, dx$$$.

Sea $$$u=\sqrt{1 - x}$$$.

Entonces $$$du=\left(\sqrt{1 - x}\right)^{\prime }dx = - \frac{1}{2 \sqrt{1 - x}} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{dx}{\sqrt{1 - x}} = - 2 du$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-2$$$ y $$$f{\left(u \right)} = \frac{u^{2}}{1 - u^{2}}$$$:

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

Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):

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

Integra término a término:

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

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

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

Realizar la descomposición en fracciones parciales (los pasos pueden verse »):

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

Integra término a término:

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{u + 1}$$$:

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

Sea $$$v=u + 1$$$.

Entonces $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dv$$$.

Por lo tanto,

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

La integral de $$$\frac{1}{v}$$$ es $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Recordemos que $$$v=u + 1$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{u - 1}$$$:

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

Sea $$$v=u - 1$$$.

Entonces $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dv$$$.

Por lo tanto,

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

La integral de $$$\frac{1}{v}$$$ es $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Recordemos que $$$v=u - 1$$$:

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

Recordemos que $$$u=\sqrt{1 - x}$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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