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

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

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

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

La integral puede reescribirse como

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

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

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

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

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

Integra término a término:

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

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

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

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

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

La integral puede reescribirse como

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

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

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

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

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

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

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

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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